mirror of https://github.com/doocs/leetcode.git
feat: update lc problems (#2506)
This commit is contained in:
Libin YANG
GitHub
parent
9b70ced919
commit
d5441b5c1e
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@ -50,7 +50,7 @@
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记 $m$ 表示字符串 $s$ 的长度，$n$ 表示字符串 $t$ 的长度。我们可以假定 $m$ 恒大于等于 $n$。
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若 $m-n\gt1$，直接返回 false；
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若 $m-n \gt 1$，直接返回 false；
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否则，遍历 $s$ 和 $t$，若遇到 $s[i]$ 不等于 $t[i]$：
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@ -59,7 +59,7 @@
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遍历结束，说明遍历过的 $s$ 跟 $t$ 所有字符相等，此时需要满足 $m=n+1$。
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时间复杂度 $O(m)$，空间复杂度 $O(1)$。
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时间复杂度 $O(m)$，其中 $m$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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@ -139,6 +139,24 @@ func isOneEditDistance(s string, t string) bool {
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}
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```
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```ts
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function isOneEditDistance(s: string, t: string): boolean {
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const [m, n] = [s.length, t.length];
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if (m < n) {
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return isOneEditDistance(t, s);
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}
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if (m - n > 1) {
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return false;
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}
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for (let i = 0; i < n; ++i) {
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if (s[i] !== t[i]) {
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return s.slice(i + 1) === t.slice(i + (m === n ? 1 : 0));
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}
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}
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return m === n + 1;
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}
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```
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<!-- tabs:end -->
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<!-- end -->
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@ -43,7 +43,20 @@
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## Solutions
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### Solution 1
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### Solution 1: Discuss Different Cases
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Let $m$ represent the length of string $s$, and $n$ represent the length of string $t$. We can assume that $m$ is always greater than or equal to $n$.
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If $m-n > 1$, return false directly;
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Otherwise, iterate through $s$ and $t$, if $s[i]$ is not equal to $t[i]$:
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- If $m \neq n$, compare $s[i+1:]$ with $t[i:]$, return true if they are equal, otherwise return false;
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- If $m = n$, compare $s[i:]$ with $t[i:]$, return true if they are equal, otherwise return false.
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If the iteration ends, it means that all the characters of $s$ and $t$ that have been iterated are equal, at this time it needs to satisfy $m=n+1$.
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The time complexity is $O(m)$, where $m$ is the length of string $s$. The space complexity is $O(1)$.
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<!-- tabs:start -->
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@ -123,6 +136,24 @@ func isOneEditDistance(s string, t string) bool {
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}
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```
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```ts
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function isOneEditDistance(s: string, t: string): boolean {
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const [m, n] = [s.length, t.length];
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if (m < n) {
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return isOneEditDistance(t, s);
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}
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if (m - n > 1) {
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return false;
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}
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for (let i = 0; i < n; ++i) {
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if (s[i] !== t[i]) {
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return s.slice(i + 1) === t.slice(i + (m === n ? 1 : 0));
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}
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}
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return m === n + 1;
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}
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```
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<!-- tabs:end -->
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<!-- end -->
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@ -0,0 +1,15 @@
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function isOneEditDistance(s: string, t: string): boolean {
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const [m, n] = [s.length, t.length];
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if (m < n) {
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return isOneEditDistance(t, s);
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}
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if (m - n > 1) {
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return false;
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}
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for (let i = 0; i < n; ++i) {
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if (s[i] !== t[i]) {
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return s.slice(i + 1) === t.slice(i + (m === n ? 1 : 0));
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}
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}
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return m === n + 1;
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}
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@ -55,7 +55,7 @@ $$
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遍历数组结束之后，每个非空的桶内的最小值和最大值都可以确定。按照桶的编号从小到大的顺序依次遍历每个桶，当前的桶的最小值和上一个非空的桶的最大值是排序后的相邻元素，计算两个相邻元素之差，并更新最大间距。遍历桶结束之后即可得到最大间距。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $nums$ 的长度。
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<!-- tabs:start -->
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@ -37,7 +37,20 @@
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## Solutions
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### Solution 1
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### Solution 1: Discuss Different Cases
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Let $m$ represent the length of string $s$, and $n$ represent the length of string $t$. We can assume that $m$ is always greater than or equal to $n$.
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If $m-n > 1$, return false directly;
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Otherwise, iterate through $s$ and $t$, if $s[i]$ is not equal to $t[i]$:
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- If $m \neq n$, compare $s[i+1:]$ with $t[i:]$, return true if they are equal, otherwise return false;
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- If $m = n$, compare $s[i:]$ with $t[i:]$, return true if they are equal, otherwise return false.
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If the iteration ends, it means that all the characters of $s$ and $t$ that have been iterated are equal, at this time it needs to satisfy $m=n+1$.
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The time complexity is $O(m)$, where $m$ is the length of string $s$. The space complexity is $O(1)$.
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<!-- tabs:start -->
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@ -159,13 +159,16 @@ func compareVersion(version1 string, version2 string) int {
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```ts
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function compareVersion(version1: string, version2: string): number {
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let v1 = version1.split('.'),
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v2 = version2.split('.');
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for (let i = 0; i < Math.max(v1.length, v2.length); i++) {
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let c1 = Number(v1[i] || 0),
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c2 = Number(v2[i] || 0);
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if (c1 > c2) return 1;
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if (c1 < c2) return -1;
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const v1 = version1.split('.');
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const v2 = version2.split('.');
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for (let i = 0; i < Math.max(v1.length, v2.length); ++i) {
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const [n1, n2] = [+v1[i] || 0, +v2[i] || 0];
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if (n1 < n2) {
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return -1;
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}
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if (n1 > n2) {
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return 1;
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}
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}
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return 0;
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}
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@ -152,13 +152,16 @@ func compareVersion(version1 string, version2 string) int {
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```ts
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function compareVersion(version1: string, version2: string): number {
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let v1 = version1.split('.'),
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v2 = version2.split('.');
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for (let i = 0; i < Math.max(v1.length, v2.length); i++) {
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let c1 = Number(v1[i] || 0),
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c2 = Number(v2[i] || 0);
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if (c1 > c2) return 1;
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if (c1 < c2) return -1;
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const v1 = version1.split('.');
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const v2 = version2.split('.');
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for (let i = 0; i < Math.max(v1.length, v2.length); ++i) {
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const [n1, n2] = [+v1[i] || 0, +v2[i] || 0];
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if (n1 < n2) {
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return -1;
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}
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if (n1 > n2) {
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return 1;
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}
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}
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return 0;
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}
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@ -1,11 +1,14 @@
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function compareVersion(version1: string, version2: string): number {
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let v1 = version1.split('.'),
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v2 = version2.split('.');
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for (let i = 0; i < Math.max(v1.length, v2.length); i++) {
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let c1 = Number(v1[i] || 0),
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c2 = Number(v2[i] || 0);
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if (c1 > c2) return 1;
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if (c1 < c2) return -1;
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const v1 = version1.split('.');
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const v2 = version2.split('.');
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for (let i = 0; i < Math.max(v1.length, v2.length); ++i) {
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const [n1, n2] = [+v1[i] || 0, +v2[i] || 0];
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if (n1 < n2) {
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return -1;
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}
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if (n1 > n2) {
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return 1;
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}
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}
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return 0;
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}
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@ -50,7 +50,21 @@
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## 解法
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### 方法一
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### 方法一：数学 + 哈希表
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我们首先判断 $numerator$ 是否为 $0$，如果是，则直接返回 `"0"`。
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接着我们判断 $numerator$ 和 $denominator$ 是否异号，如果是，则结果为负数，我们将结果的第一个字符设为 `"-"`。
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然后我们将 $numerator$ 和 $denominator$ 取绝对值，分别记为 $a$ 和 $b$。由于分子和分母的范围为 $[-2^{31}, 2^{31} - 1]$，直接取绝对值可能会溢出，因此我们将 $a$ 和 $b$ 都转换为长整型。
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接着我们计算整数部分，即 $a$ 除以 $b$ 的整数部分，将其转换为字符串并添加到结果中。然后我们将 $a$ 取余 $b$，记为 $a$。
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如果 $a$ 为 $0$，说明结果为整数，直接返回结果。
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接着我们计算小数部分，我们使用哈希表 $d$ 记录每个余数对应的结果的长度。我们不断将 $a$ 乘以 $10$，然后将 $a$ 除以 $b$ 的整数部分添加到结果中，然后将 $a$ 取余 $b$，记为 $a$。如果 $a$ 为 $0$，说明结果为有限小数，直接返回结果。如果 $a$ 在哈希表中出现过，说明结果为循环小数，我们找到循环的起始位置，将结果插入括号中，然后返回结果。
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时间复杂度 $O(l)$，空间复杂度 $O(l)$，其中 $l$ 为结果的长度，本题中 $l < 10^4$。
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<!-- tabs:start -->
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@ -58,29 +72,28 @@
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class Solution:
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def fractionToDecimal(self, numerator: int, denominator: int) -> str:
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if numerator == 0:
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return '0'
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res = []
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return "0"
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ans = []
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neg = (numerator > 0) ^ (denominator > 0)
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if neg:
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res.append('-')
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num, d = abs(numerator), abs(denominator)
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res.append(str(num // d))
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num %= d
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if num == 0:
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return ''.join(res)
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res.append('.')
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mp = {}
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while num != 0:
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mp[num] = len(res)
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num *= 10
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res.append(str(num // d))
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num %= d
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if num in mp:
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idx = mp[num]
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res.insert(idx, '(')
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res.append(')')
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ans.append("-")
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a, b = abs(numerator), abs(denominator)
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ans.append(str(a // b))
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a %= b
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if a == 0:
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return "".join(ans)
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ans.append(".")
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d = {}
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while a:
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d[a] = len(ans)
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a *= 10
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ans.append(str(a // b))
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a %= b
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if a in d:
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ans.insert(d[a], "(")
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ans.append(")")
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break
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return ''.join(res)
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return "".join(ans)
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```
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```java
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@ -92,23 +105,21 @@ class Solution {
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StringBuilder sb = new StringBuilder();
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boolean neg = (numerator > 0) ^ (denominator > 0);
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sb.append(neg ? "-" : "");
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long num = Math.abs((long) numerator);
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long d = Math.abs((long) denominator);
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sb.append(num / d);
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num %= d;
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if (num == 0) {
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long a = Math.abs((long) numerator), b = Math.abs((long) denominator);
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sb.append(a / b);
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a %= b;
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if (a == 0) {
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return sb.toString();
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}
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sb.append(".");
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Map<Long, Integer> mp = new HashMap<>();
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while (num != 0) {
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mp.put(num, sb.length());
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num *= 10;
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sb.append(num / d);
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num %= d;
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if (mp.containsKey(num)) {
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int idx = mp.get(num);
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sb.insert(idx, "(");
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Map<Long, Integer> d = new HashMap<>();
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while (a != 0) {
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d.put(a, sb.length());
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a *= 10;
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sb.append(a / b);
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a %= b;
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if (d.containsKey(a)) {
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sb.insert(d.get(a), "(");
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sb.append(")");
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break;
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}
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@ -119,35 +130,37 @@ class Solution {
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```
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```cpp
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using LL = long long;
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class Solution {
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public:
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string fractionToDecimal(int numerator, int denominator) {
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if (numerator == 0) return "0";
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string res = "";
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if (numerator == 0) {
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return "0";
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}
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string ans;
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bool neg = (numerator > 0) ^ (denominator > 0);
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if (neg) res += "-";
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LL num = abs(numerator);
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LL d = abs(denominator);
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res += to_string(num / d);
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num %= d;
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if (num == 0) return res;
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res += ".";
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unordered_map<LL, int> mp;
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while (num) {
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mp[num] = res.size();
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num *= 10;
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res += to_string(num / d);
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num %= d;
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if (mp.count(num)) {
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int idx = mp[num];
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res.insert(idx, "(");
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res += ")";
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if (neg) {
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ans += "-";
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}
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long long a = abs(1LL * numerator), b = abs(1LL * denominator);
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ans += to_string(a / b);
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a %= b;
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if (a == 0) {
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return ans;
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}
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ans += ".";
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unordered_map<long long, int> d;
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while (a) {
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d[a] = ans.size();
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a *= 10;
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ans += to_string(a / b);
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a %= b;
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if (d.contains(a)) {
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ans.insert(d[a], "(");
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ans += ")";
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break;
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}
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}
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return res;
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return ans;
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}
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};
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```
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@ -157,37 +170,35 @@ func fractionToDecimal(numerator int, denominator int) string {
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if numerator == 0 {
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return "0"
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}
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res := []byte{}
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neg := numerator*denominator < 0
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if neg {
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res = append(res, '-')
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ans := ""
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if (numerator > 0) != (denominator > 0) {
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ans += "-"
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}
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num := abs(numerator)
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d := abs(denominator)
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res = append(res, strconv.Itoa(num/d)...)
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num %= d
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if num == 0 {
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return string(res)
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a := int64(numerator)
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b := int64(denominator)
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a = abs(a)
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b = abs(b)
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ans += strconv.FormatInt(a/b, 10)
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a %= b
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if a == 0 {
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return ans
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}
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mp := make(map[int]int)
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res = append(res, '.')
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for num != 0 {
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mp[num] = len(res)
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num *= 10
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res = append(res, strconv.Itoa(num/d)...)
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num %= d
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if mp[num] > 0 {
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idx := mp[num]
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res = append(res[:idx], append([]byte{'('}, res[idx:]...)...)
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res = append(res, ')')
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ans += "."
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d := make(map[int64]int)
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for a != 0 {
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if pos, ok := d[a]; ok {
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ans = ans[:pos] + "(" + ans[pos:] + ")"
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break
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}
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d[a] = len(ans)
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a *= 10
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ans += strconv.FormatInt(a/b, 10)
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a %= b
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}
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return string(res)
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return ans
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}
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func abs(x int) int {
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func abs(x int64) int64 {
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if x < 0 {
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return -x
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||||
}
|
||||
|
|
@ -195,59 +206,64 @@ func abs(x int) int {
|
|||
}
|
||||
```
|
||||
|
||||
```ts
|
||||
function fractionToDecimal(numerator: number, denominator: number): string {
|
||||
if (numerator === 0) {
|
||||
return '0';
|
||||
}
|
||||
const sb: string[] = [];
|
||||
const neg: boolean = numerator > 0 !== denominator > 0;
|
||||
sb.push(neg ? '-' : '');
|
||||
let a: number = Math.abs(numerator),
|
||||
b: number = Math.abs(denominator);
|
||||
sb.push(Math.floor(a / b).toString());
|
||||
a %= b;
|
||||
if (a === 0) {
|
||||
return sb.join('');
|
||||
}
|
||||
sb.push('.');
|
||||
const d: Map<number, number> = new Map();
|
||||
while (a !== 0) {
|
||||
d.set(a, sb.length);
|
||||
a *= 10;
|
||||
sb.push(Math.floor(a / b).toString());
|
||||
a %= b;
|
||||
if (d.has(a)) {
|
||||
sb.splice(d.get(a)!, 0, '(');
|
||||
sb.push(')');
|
||||
break;
|
||||
}
|
||||
}
|
||||
return sb.join('');
|
||||
}
|
||||
```
|
||||
|
||||
```cs
|
||||
// https://leetcode.com/problems/fraction-to-recurring-decimal/
|
||||
|
||||
using System.Collections.Generic;
|
||||
using System.Text;
|
||||
|
||||
public partial class Solution
|
||||
{
|
||||
public string FractionToDecimal(int numerator, int denominator)
|
||||
{
|
||||
var n = (long)numerator;
|
||||
var d = (long)denominator;
|
||||
var sb = new StringBuilder();
|
||||
if (n < 0)
|
||||
{
|
||||
n = -n;
|
||||
if (d < 0)
|
||||
{
|
||||
d = -d;
|
||||
}
|
||||
else
|
||||
{
|
||||
sb.Append('-');
|
||||
}
|
||||
public class Solution {
|
||||
public string FractionToDecimal(int numerator, int denominator) {
|
||||
if (numerator == 0) {
|
||||
return "0";
|
||||
}
|
||||
else if (n > 0 && d < 0)
|
||||
{
|
||||
d = -d;
|
||||
sb.Append('-');
|
||||
StringBuilder sb = new StringBuilder();
|
||||
bool neg = (numerator > 0) ^ (denominator > 0);
|
||||
sb.Append(neg ? "-" : "");
|
||||
long a = Math.Abs((long)numerator), b = Math.Abs((long)denominator);
|
||||
sb.Append(a / b);
|
||||
a %= b;
|
||||
if (a == 0) {
|
||||
return sb.ToString();
|
||||
}
|
||||
|
||||
sb.Append(n / d);
|
||||
n = n % d;
|
||||
if (n != 0)
|
||||
{
|
||||
sb.Append('.');
|
||||
var dict = new Dictionary<long, int>();
|
||||
while (n != 0)
|
||||
{
|
||||
int index;
|
||||
if (dict.TryGetValue(n, out index))
|
||||
{
|
||||
sb.Insert(index, '(');
|
||||
sb.Append(')');
|
||||
break;
|
||||
}
|
||||
else
|
||||
{
|
||||
dict.Add(n, sb.Length);
|
||||
n *= 10;
|
||||
sb.Append(n / d);
|
||||
n %= d;
|
||||
}
|
||||
sb.Append(".");
|
||||
Dictionary<long, int> d = new Dictionary<long, int>();
|
||||
while (a != 0) {
|
||||
d[a] = sb.Length;
|
||||
a *= 10;
|
||||
sb.Append(a / b);
|
||||
a %= b;
|
||||
if (d.ContainsKey(a)) {
|
||||
sb.Insert(d[a], "(");
|
||||
sb.Append(")");
|
||||
break;
|
||||
}
|
||||
}
|
||||
return sb.ToString();
|
||||
|
|
|
|||
|
|
@ -46,7 +46,21 @@
|
|||
|
||||
## Solutions
|
||||
|
||||
### Solution 1
|
||||
### Solution 1: Mathematics + Hash Table
|
||||
|
||||
First, we check if the $numerator$ is $0$. If it is, we return `"0"` directly.
|
||||
|
||||
Next, we check if the $numerator$ and $denominator$ have different signs. If they do, the result is negative, and we set the first character of the result to `"-"`.
|
||||
|
||||
Then we take the absolute values of the $numerator$ and $denominator$, denoted as $a$ and $b$. Since the range of the numerator and denominator is $[-2^{31}, 2^{31} - 1]$, taking the absolute value directly may cause overflow, so we convert both $a$ and $b$ to long integers.
|
||||
|
||||
Next, we calculate the integer part, which is the integer part of $a$ divided by $b$, convert it to a string, and add it to the result. Then we take the remainder of $a$ divided by $b$, denoted as $a$.
|
||||
|
||||
If $a$ is $0$, it means the result is an integer, and we return the result directly.
|
||||
|
||||
Next, we calculate the decimal part. We use a hash table $d$ to record the length of the result corresponding to each remainder. We continuously multiply $a$ by $10$, then add the integer part of $a$ divided by $b$ to the result, then take the remainder of $a$ divided by $b$, denoted as $a$. If $a$ is $0$, it means the result is a finite decimal, and we return the result directly. If $a$ has appeared in the hash table, it means the result is a recurring decimal. We find the starting position of the cycle, insert the result into parentheses, and then return the result.
|
||||
|
||||
The time complexity is $O(l)$, and the space complexity is $O(l)$, where $l$ is the length of the result. In this problem, $l < 10^4$.
|
||||
|
||||
<!-- tabs:start -->
|
||||
|
||||
|
|
@ -54,29 +68,28 @@
|
|||
class Solution:
|
||||
def fractionToDecimal(self, numerator: int, denominator: int) -> str:
|
||||
if numerator == 0:
|
||||
return '0'
|
||||
res = []
|
||||
return "0"
|
||||
ans = []
|
||||
neg = (numerator > 0) ^ (denominator > 0)
|
||||
if neg:
|
||||
res.append('-')
|
||||
num, d = abs(numerator), abs(denominator)
|
||||
res.append(str(num // d))
|
||||
num %= d
|
||||
if num == 0:
|
||||
return ''.join(res)
|
||||
res.append('.')
|
||||
mp = {}
|
||||
while num != 0:
|
||||
mp[num] = len(res)
|
||||
num *= 10
|
||||
res.append(str(num // d))
|
||||
num %= d
|
||||
if num in mp:
|
||||
idx = mp[num]
|
||||
res.insert(idx, '(')
|
||||
res.append(')')
|
||||
ans.append("-")
|
||||
a, b = abs(numerator), abs(denominator)
|
||||
ans.append(str(a // b))
|
||||
a %= b
|
||||
if a == 0:
|
||||
return "".join(ans)
|
||||
ans.append(".")
|
||||
d = {}
|
||||
while a:
|
||||
d[a] = len(ans)
|
||||
a *= 10
|
||||
ans.append(str(a // b))
|
||||
a %= b
|
||||
if a in d:
|
||||
ans.insert(d[a], "(")
|
||||
ans.append(")")
|
||||
break
|
||||
return ''.join(res)
|
||||
return "".join(ans)
|
||||
```
|
||||
|
||||
```java
|
||||
|
|
@ -88,23 +101,21 @@ class Solution {
|
|||
StringBuilder sb = new StringBuilder();
|
||||
boolean neg = (numerator > 0) ^ (denominator > 0);
|
||||
sb.append(neg ? "-" : "");
|
||||
long num = Math.abs((long) numerator);
|
||||
long d = Math.abs((long) denominator);
|
||||
sb.append(num / d);
|
||||
num %= d;
|
||||
if (num == 0) {
|
||||
long a = Math.abs((long) numerator), b = Math.abs((long) denominator);
|
||||
sb.append(a / b);
|
||||
a %= b;
|
||||
if (a == 0) {
|
||||
return sb.toString();
|
||||
}
|
||||
sb.append(".");
|
||||
Map<Long, Integer> mp = new HashMap<>();
|
||||
while (num != 0) {
|
||||
mp.put(num, sb.length());
|
||||
num *= 10;
|
||||
sb.append(num / d);
|
||||
num %= d;
|
||||
if (mp.containsKey(num)) {
|
||||
int idx = mp.get(num);
|
||||
sb.insert(idx, "(");
|
||||
Map<Long, Integer> d = new HashMap<>();
|
||||
while (a != 0) {
|
||||
d.put(a, sb.length());
|
||||
a *= 10;
|
||||
sb.append(a / b);
|
||||
a %= b;
|
||||
if (d.containsKey(a)) {
|
||||
sb.insert(d.get(a), "(");
|
||||
sb.append(")");
|
||||
break;
|
||||
}
|
||||
|
|
@ -115,35 +126,37 @@ class Solution {
|
|||
```
|
||||
|
||||
```cpp
|
||||
using LL = long long;
|
||||
|
||||
class Solution {
|
||||
public:
|
||||
string fractionToDecimal(int numerator, int denominator) {
|
||||
if (numerator == 0) return "0";
|
||||
string res = "";
|
||||
if (numerator == 0) {
|
||||
return "0";
|
||||
}
|
||||
string ans;
|
||||
bool neg = (numerator > 0) ^ (denominator > 0);
|
||||
if (neg) res += "-";
|
||||
LL num = abs(numerator);
|
||||
LL d = abs(denominator);
|
||||
res += to_string(num / d);
|
||||
num %= d;
|
||||
if (num == 0) return res;
|
||||
res += ".";
|
||||
unordered_map<LL, int> mp;
|
||||
while (num) {
|
||||
mp[num] = res.size();
|
||||
num *= 10;
|
||||
res += to_string(num / d);
|
||||
num %= d;
|
||||
if (mp.count(num)) {
|
||||
int idx = mp[num];
|
||||
res.insert(idx, "(");
|
||||
res += ")";
|
||||
if (neg) {
|
||||
ans += "-";
|
||||
}
|
||||
long long a = abs(1LL * numerator), b = abs(1LL * denominator);
|
||||
ans += to_string(a / b);
|
||||
a %= b;
|
||||
if (a == 0) {
|
||||
return ans;
|
||||
}
|
||||
ans += ".";
|
||||
unordered_map<long long, int> d;
|
||||
while (a) {
|
||||
d[a] = ans.size();
|
||||
a *= 10;
|
||||
ans += to_string(a / b);
|
||||
a %= b;
|
||||
if (d.contains(a)) {
|
||||
ans.insert(d[a], "(");
|
||||
ans += ")";
|
||||
break;
|
||||
}
|
||||
}
|
||||
return res;
|
||||
return ans;
|
||||
}
|
||||
};
|
||||
```
|
||||
|
|
@ -153,37 +166,35 @@ func fractionToDecimal(numerator int, denominator int) string {
|
|||
if numerator == 0 {
|
||||
return "0"
|
||||
}
|
||||
res := []byte{}
|
||||
neg := numerator*denominator < 0
|
||||
if neg {
|
||||
res = append(res, '-')
|
||||
ans := ""
|
||||
if (numerator > 0) != (denominator > 0) {
|
||||
ans += "-"
|
||||
}
|
||||
num := abs(numerator)
|
||||
d := abs(denominator)
|
||||
res = append(res, strconv.Itoa(num/d)...)
|
||||
num %= d
|
||||
if num == 0 {
|
||||
return string(res)
|
||||
a := int64(numerator)
|
||||
b := int64(denominator)
|
||||
a = abs(a)
|
||||
b = abs(b)
|
||||
ans += strconv.FormatInt(a/b, 10)
|
||||
a %= b
|
||||
if a == 0 {
|
||||
return ans
|
||||
}
|
||||
mp := make(map[int]int)
|
||||
res = append(res, '.')
|
||||
for num != 0 {
|
||||
mp[num] = len(res)
|
||||
num *= 10
|
||||
res = append(res, strconv.Itoa(num/d)...)
|
||||
num %= d
|
||||
if mp[num] > 0 {
|
||||
idx := mp[num]
|
||||
res = append(res[:idx], append([]byte{'('}, res[idx:]...)...)
|
||||
res = append(res, ')')
|
||||
ans += "."
|
||||
d := make(map[int64]int)
|
||||
for a != 0 {
|
||||
if pos, ok := d[a]; ok {
|
||||
ans = ans[:pos] + "(" + ans[pos:] + ")"
|
||||
break
|
||||
}
|
||||
d[a] = len(ans)
|
||||
a *= 10
|
||||
ans += strconv.FormatInt(a/b, 10)
|
||||
a %= b
|
||||
}
|
||||
|
||||
return string(res)
|
||||
return ans
|
||||
}
|
||||
|
||||
func abs(x int) int {
|
||||
func abs(x int64) int64 {
|
||||
if x < 0 {
|
||||
return -x
|
||||
}
|
||||
|
|
@ -191,59 +202,64 @@ func abs(x int) int {
|
|||
}
|
||||
```
|
||||
|
||||
```ts
|
||||
function fractionToDecimal(numerator: number, denominator: number): string {
|
||||
if (numerator === 0) {
|
||||
return '0';
|
||||
}
|
||||
const sb: string[] = [];
|
||||
const neg: boolean = numerator > 0 !== denominator > 0;
|
||||
sb.push(neg ? '-' : '');
|
||||
let a: number = Math.abs(numerator),
|
||||
b: number = Math.abs(denominator);
|
||||
sb.push(Math.floor(a / b).toString());
|
||||
a %= b;
|
||||
if (a === 0) {
|
||||
return sb.join('');
|
||||
}
|
||||
sb.push('.');
|
||||
const d: Map<number, number> = new Map();
|
||||
while (a !== 0) {
|
||||
d.set(a, sb.length);
|
||||
a *= 10;
|
||||
sb.push(Math.floor(a / b).toString());
|
||||
a %= b;
|
||||
if (d.has(a)) {
|
||||
sb.splice(d.get(a)!, 0, '(');
|
||||
sb.push(')');
|
||||
break;
|
||||
}
|
||||
}
|
||||
return sb.join('');
|
||||
}
|
||||
```
|
||||
|
||||
```cs
|
||||
// https://leetcode.com/problems/fraction-to-recurring-decimal/
|
||||
|
||||
using System.Collections.Generic;
|
||||
using System.Text;
|
||||
|
||||
public partial class Solution
|
||||
{
|
||||
public string FractionToDecimal(int numerator, int denominator)
|
||||
{
|
||||
var n = (long)numerator;
|
||||
var d = (long)denominator;
|
||||
var sb = new StringBuilder();
|
||||
if (n < 0)
|
||||
{
|
||||
n = -n;
|
||||
if (d < 0)
|
||||
{
|
||||
d = -d;
|
||||
}
|
||||
else
|
||||
{
|
||||
sb.Append('-');
|
||||
}
|
||||
public class Solution {
|
||||
public string FractionToDecimal(int numerator, int denominator) {
|
||||
if (numerator == 0) {
|
||||
return "0";
|
||||
}
|
||||
else if (n > 0 && d < 0)
|
||||
{
|
||||
d = -d;
|
||||
sb.Append('-');
|
||||
StringBuilder sb = new StringBuilder();
|
||||
bool neg = (numerator > 0) ^ (denominator > 0);
|
||||
sb.Append(neg ? "-" : "");
|
||||
long a = Math.Abs((long)numerator), b = Math.Abs((long)denominator);
|
||||
sb.Append(a / b);
|
||||
a %= b;
|
||||
if (a == 0) {
|
||||
return sb.ToString();
|
||||
}
|
||||
|
||||
sb.Append(n / d);
|
||||
n = n % d;
|
||||
if (n != 0)
|
||||
{
|
||||
sb.Append('.');
|
||||
var dict = new Dictionary<long, int>();
|
||||
while (n != 0)
|
||||
{
|
||||
int index;
|
||||
if (dict.TryGetValue(n, out index))
|
||||
{
|
||||
sb.Insert(index, '(');
|
||||
sb.Append(')');
|
||||
break;
|
||||
}
|
||||
else
|
||||
{
|
||||
dict.Add(n, sb.Length);
|
||||
n *= 10;
|
||||
sb.Append(n / d);
|
||||
n %= d;
|
||||
}
|
||||
sb.Append(".");
|
||||
Dictionary<long, int> d = new Dictionary<long, int>();
|
||||
while (a != 0) {
|
||||
d[a] = sb.Length;
|
||||
a *= 10;
|
||||
sb.Append(a / b);
|
||||
a %= b;
|
||||
if (d.ContainsKey(a)) {
|
||||
sb.Insert(d[a], "(");
|
||||
sb.Append(")");
|
||||
break;
|
||||
}
|
||||
}
|
||||
return sb.ToString();
|
||||
|
|
|
|||
|
|
@ -1,31 +1,33 @@
|
|||
using LL = long long;
|
||||
|
||||
class Solution {
|
||||
public:
|
||||
string fractionToDecimal(int numerator, int denominator) {
|
||||
if (numerator == 0) return "0";
|
||||
string res = "";
|
||||
if (numerator == 0) {
|
||||
return "0";
|
||||
}
|
||||
string ans;
|
||||
bool neg = (numerator > 0) ^ (denominator > 0);
|
||||
if (neg) res += "-";
|
||||
LL num = abs(numerator);
|
||||
LL d = abs(denominator);
|
||||
res += to_string(num / d);
|
||||
num %= d;
|
||||
if (num == 0) return res;
|
||||
res += ".";
|
||||
unordered_map<LL, int> mp;
|
||||
while (num) {
|
||||
mp[num] = res.size();
|
||||
num *= 10;
|
||||
res += to_string(num / d);
|
||||
num %= d;
|
||||
if (mp.count(num)) {
|
||||
int idx = mp[num];
|
||||
res.insert(idx, "(");
|
||||
res += ")";
|
||||
if (neg) {
|
||||
ans += "-";
|
||||
}
|
||||
long long a = abs(1LL * numerator), b = abs(1LL * denominator);
|
||||
ans += to_string(a / b);
|
||||
a %= b;
|
||||
if (a == 0) {
|
||||
return ans;
|
||||
}
|
||||
ans += ".";
|
||||
unordered_map<long long, int> d;
|
||||
while (a) {
|
||||
d[a] = ans.size();
|
||||
a *= 10;
|
||||
ans += to_string(a / b);
|
||||
a %= b;
|
||||
if (d.contains(a)) {
|
||||
ans.insert(d[a], "(");
|
||||
ans += ")";
|
||||
break;
|
||||
}
|
||||
}
|
||||
return res;
|
||||
return ans;
|
||||
}
|
||||
};
|
||||
|
|
@ -1,55 +1,28 @@
|
|||
// https://leetcode.com/problems/fraction-to-recurring-decimal/
|
||||
|
||||
using System.Collections.Generic;
|
||||
using System.Text;
|
||||
|
||||
public partial class Solution
|
||||
{
|
||||
public string FractionToDecimal(int numerator, int denominator)
|
||||
{
|
||||
var n = (long)numerator;
|
||||
var d = (long)denominator;
|
||||
var sb = new StringBuilder();
|
||||
if (n < 0)
|
||||
{
|
||||
n = -n;
|
||||
if (d < 0)
|
||||
{
|
||||
d = -d;
|
||||
}
|
||||
else
|
||||
{
|
||||
sb.Append('-');
|
||||
}
|
||||
public class Solution {
|
||||
public string FractionToDecimal(int numerator, int denominator) {
|
||||
if (numerator == 0) {
|
||||
return "0";
|
||||
}
|
||||
else if (n > 0 && d < 0)
|
||||
{
|
||||
d = -d;
|
||||
sb.Append('-');
|
||||
StringBuilder sb = new StringBuilder();
|
||||
bool neg = (numerator > 0) ^ (denominator > 0);
|
||||
sb.Append(neg ? "-" : "");
|
||||
long a = Math.Abs((long)numerator), b = Math.Abs((long)denominator);
|
||||
sb.Append(a / b);
|
||||
a %= b;
|
||||
if (a == 0) {
|
||||
return sb.ToString();
|
||||
}
|
||||
|
||||
sb.Append(n / d);
|
||||
n = n % d;
|
||||
if (n != 0)
|
||||
{
|
||||
sb.Append('.');
|
||||
var dict = new Dictionary<long, int>();
|
||||
while (n != 0)
|
||||
{
|
||||
int index;
|
||||
if (dict.TryGetValue(n, out index))
|
||||
{
|
||||
sb.Insert(index, '(');
|
||||
sb.Append(')');
|
||||
break;
|
||||
}
|
||||
else
|
||||
{
|
||||
dict.Add(n, sb.Length);
|
||||
n *= 10;
|
||||
sb.Append(n / d);
|
||||
n %= d;
|
||||
}
|
||||
sb.Append(".");
|
||||
Dictionary<long, int> d = new Dictionary<long, int>();
|
||||
while (a != 0) {
|
||||
d[a] = sb.Length;
|
||||
a *= 10;
|
||||
sb.Append(a / b);
|
||||
a %= b;
|
||||
if (d.ContainsKey(a)) {
|
||||
sb.Insert(d[a], "(");
|
||||
sb.Append(")");
|
||||
break;
|
||||
}
|
||||
}
|
||||
return sb.ToString();
|
||||
|
|
|
|||
|
|
@ -2,37 +2,35 @@ func fractionToDecimal(numerator int, denominator int) string {
|
|||
if numerator == 0 {
|
||||
return "0"
|
||||
}
|
||||
res := []byte{}
|
||||
neg := numerator*denominator < 0
|
||||
if neg {
|
||||
res = append(res, '-')
|
||||
ans := ""
|
||||
if (numerator > 0) != (denominator > 0) {
|
||||
ans += "-"
|
||||
}
|
||||
num := abs(numerator)
|
||||
d := abs(denominator)
|
||||
res = append(res, strconv.Itoa(num/d)...)
|
||||
num %= d
|
||||
if num == 0 {
|
||||
return string(res)
|
||||
a := int64(numerator)
|
||||
b := int64(denominator)
|
||||
a = abs(a)
|
||||
b = abs(b)
|
||||
ans += strconv.FormatInt(a/b, 10)
|
||||
a %= b
|
||||
if a == 0 {
|
||||
return ans
|
||||
}
|
||||
mp := make(map[int]int)
|
||||
res = append(res, '.')
|
||||
for num != 0 {
|
||||
mp[num] = len(res)
|
||||
num *= 10
|
||||
res = append(res, strconv.Itoa(num/d)...)
|
||||
num %= d
|
||||
if mp[num] > 0 {
|
||||
idx := mp[num]
|
||||
res = append(res[:idx], append([]byte{'('}, res[idx:]...)...)
|
||||
res = append(res, ')')
|
||||
ans += "."
|
||||
d := make(map[int64]int)
|
||||
for a != 0 {
|
||||
if pos, ok := d[a]; ok {
|
||||
ans = ans[:pos] + "(" + ans[pos:] + ")"
|
||||
break
|
||||
}
|
||||
d[a] = len(ans)
|
||||
a *= 10
|
||||
ans += strconv.FormatInt(a/b, 10)
|
||||
a %= b
|
||||
}
|
||||
|
||||
return string(res)
|
||||
return ans
|
||||
}
|
||||
|
||||
func abs(x int) int {
|
||||
func abs(x int64) int64 {
|
||||
if x < 0 {
|
||||
return -x
|
||||
}
|
||||
|
|
|
|||
|
|
@ -6,23 +6,21 @@ class Solution {
|
|||
StringBuilder sb = new StringBuilder();
|
||||
boolean neg = (numerator > 0) ^ (denominator > 0);
|
||||
sb.append(neg ? "-" : "");
|
||||
long num = Math.abs((long) numerator);
|
||||
long d = Math.abs((long) denominator);
|
||||
sb.append(num / d);
|
||||
num %= d;
|
||||
if (num == 0) {
|
||||
long a = Math.abs((long) numerator), b = Math.abs((long) denominator);
|
||||
sb.append(a / b);
|
||||
a %= b;
|
||||
if (a == 0) {
|
||||
return sb.toString();
|
||||
}
|
||||
sb.append(".");
|
||||
Map<Long, Integer> mp = new HashMap<>();
|
||||
while (num != 0) {
|
||||
mp.put(num, sb.length());
|
||||
num *= 10;
|
||||
sb.append(num / d);
|
||||
num %= d;
|
||||
if (mp.containsKey(num)) {
|
||||
int idx = mp.get(num);
|
||||
sb.insert(idx, "(");
|
||||
Map<Long, Integer> d = new HashMap<>();
|
||||
while (a != 0) {
|
||||
d.put(a, sb.length());
|
||||
a *= 10;
|
||||
sb.append(a / b);
|
||||
a %= b;
|
||||
if (d.containsKey(a)) {
|
||||
sb.insert(d.get(a), "(");
|
||||
sb.append(")");
|
||||
break;
|
||||
}
|
||||
|
|
|
|||
|
|
@ -1,26 +1,25 @@
|
|||
class Solution:
|
||||
def fractionToDecimal(self, numerator: int, denominator: int) -> str:
|
||||
if numerator == 0:
|
||||
return '0'
|
||||
res = []
|
||||
return "0"
|
||||
ans = []
|
||||
neg = (numerator > 0) ^ (denominator > 0)
|
||||
if neg:
|
||||
res.append('-')
|
||||
num, d = abs(numerator), abs(denominator)
|
||||
res.append(str(num // d))
|
||||
num %= d
|
||||
if num == 0:
|
||||
return ''.join(res)
|
||||
res.append('.')
|
||||
mp = {}
|
||||
while num != 0:
|
||||
mp[num] = len(res)
|
||||
num *= 10
|
||||
res.append(str(num // d))
|
||||
num %= d
|
||||
if num in mp:
|
||||
idx = mp[num]
|
||||
res.insert(idx, '(')
|
||||
res.append(')')
|
||||
ans.append("-")
|
||||
a, b = abs(numerator), abs(denominator)
|
||||
ans.append(str(a // b))
|
||||
a %= b
|
||||
if a == 0:
|
||||
return "".join(ans)
|
||||
ans.append(".")
|
||||
d = {}
|
||||
while a:
|
||||
d[a] = len(ans)
|
||||
a *= 10
|
||||
ans.append(str(a // b))
|
||||
a %= b
|
||||
if a in d:
|
||||
ans.insert(d[a], "(")
|
||||
ans.append(")")
|
||||
break
|
||||
return ''.join(res)
|
||||
return "".join(ans)
|
||||
|
|
|
|||
|
|
@ -0,0 +1,29 @@
|
|||
function fractionToDecimal(numerator: number, denominator: number): string {
|
||||
if (numerator === 0) {
|
||||
return '0';
|
||||
}
|
||||
const sb: string[] = [];
|
||||
const neg: boolean = numerator > 0 !== denominator > 0;
|
||||
sb.push(neg ? '-' : '');
|
||||
let a: number = Math.abs(numerator),
|
||||
b: number = Math.abs(denominator);
|
||||
sb.push(Math.floor(a / b).toString());
|
||||
a %= b;
|
||||
if (a === 0) {
|
||||
return sb.join('');
|
||||
}
|
||||
sb.push('.');
|
||||
const d: Map<number, number> = new Map();
|
||||
while (a !== 0) {
|
||||
d.set(a, sb.length);
|
||||
a *= 10;
|
||||
sb.push(Math.floor(a / b).toString());
|
||||
a %= b;
|
||||
if (d.has(a)) {
|
||||
sb.splice(d.get(a)!, 0, '(');
|
||||
sb.push(')');
|
||||
break;
|
||||
}
|
||||
}
|
||||
return sb.join('');
|
||||
}
|
||||
|
|
@ -6,39 +6,44 @@
|
|||
|
||||
## Description
|
||||
|
||||
<p>Write a function that takes the binary representation of an unsigned integer and returns the number of '1' bits it has (also known as the <a href="http://en.wikipedia.org/wiki/Hamming_weight" target="_blank">Hamming weight</a>).</p>
|
||||
|
||||
<p><strong>Note:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.</li>
|
||||
<li>In Java, the compiler represents the signed integers using <a href="https://en.wikipedia.org/wiki/Two%27s_complement" target="_blank">2's complement notation</a>. Therefore, in <strong class="example">Example 3</strong>, the input represents the signed integer. <code>-3</code>.</li>
|
||||
</ul>
|
||||
<p>Write a function that takes the binary representation of a positive integer and returns the number of <span data-keyword="set-bit">set bits</span> it has (also known as the <a href="http://en.wikipedia.org/wiki/Hamming_weight" target="_blank">Hamming weight</a>).</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 00000000000000000000000000001011
|
||||
<strong>Output:</strong> 3
|
||||
<strong>Explanation:</strong> The input binary string <strong>00000000000000000000000000001011</strong> has a total of three '1' bits.
|
||||
</pre>
|
||||
<div class="example-block">
|
||||
<p><strong>Input:</strong> <span class="example-io">n = 11</span></p>
|
||||
|
||||
<p><strong>Output:</strong> <span class="example-io">3</span></p>
|
||||
|
||||
<p><strong>Explanation:</strong></p>
|
||||
|
||||
<p>The input binary string <strong>1011</strong> has a total of three set bits.</p>
|
||||
</div>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 00000000000000000000000010000000
|
||||
<strong>Output:</strong> 1
|
||||
<strong>Explanation:</strong> The input binary string <strong>00000000000000000000000010000000</strong> has a total of one '1' bit.
|
||||
</pre>
|
||||
<div class="example-block">
|
||||
<p><strong>Input:</strong> <span class="example-io">n = 128</span></p>
|
||||
|
||||
<p><strong>Output:</strong> <span class="example-io">1</span></p>
|
||||
|
||||
<p><strong>Explanation:</strong></p>
|
||||
|
||||
<p>The input binary string <strong>10000000</strong> has a total of one set bit.</p>
|
||||
</div>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 11111111111111111111111111111101
|
||||
<strong>Output:</strong> 31
|
||||
<strong>Explanation:</strong> The input binary string <strong>11111111111111111111111111111101</strong> has a total of thirty one '1' bits.
|
||||
</pre>
|
||||
<div class="example-block">
|
||||
<p><strong>Input:</strong> <span class="example-io">n = 2147483645</span></p>
|
||||
|
||||
<p><strong>Output:</strong> <span class="example-io">30</span></p>
|
||||
|
||||
<p><strong>Explanation:</strong></p>
|
||||
|
||||
<p>The input binary string <strong>1111111111111111111111111111101</strong> has a total of thirty set bits.</p>
|
||||
</div>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[English Version](/solution/0300-0399/0321.Create%20Maximum%20Number/README_EN.md)
|
||||
|
||||
<!-- tags:栈,贪心,单调栈 -->
|
||||
<!-- tags:栈,贪心,数组,双指针,单调栈 -->
|
||||
|
||||
## 题目描述
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[中文文档](/solution/0300-0399/0321.Create%20Maximum%20Number/README.md)
|
||||
|
||||
<!-- tags:Stack,Greedy,Monotonic Stack -->
|
||||
<!-- tags:Stack,Greedy,Array,Two Pointers,Monotonic Stack -->
|
||||
|
||||
## Description
|
||||
|
||||
|
|
|
|||
|
|
@ -8,7 +8,7 @@
|
|||
|
||||
<!-- 这里写题目描述 -->
|
||||
|
||||
<p>给定两个数组 <code>nums1</code> 和 <code>nums2</code> ，返回 <em>它们的交集</em> 。输出结果中的每个元素一定是 <strong>唯一</strong> 的。我们可以 <strong>不考虑输出结果的顺序</strong> 。</p>
|
||||
<p>给定两个数组 <code>nums1</code> 和 <code>nums2</code> ，返回 <em>它们的 <span data-keyword="array-intersection">交集</span></em> 。输出结果中的每个元素一定是 <strong>唯一</strong> 的。我们可以 <strong>不考虑输出结果的顺序</strong> 。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
|
|
|
|||
|
|
@ -6,7 +6,7 @@
|
|||
|
||||
## Description
|
||||
|
||||
<p>Given two integer arrays <code>nums1</code> and <code>nums2</code>, return <em>an array of their intersection</em>. Each element in the result must be <strong>unique</strong> and you may return the result in <strong>any order</strong>.</p>
|
||||
<p>Given two integer arrays <code>nums1</code> and <code>nums2</code>, return <em>an array of their <span data-keyword="array-intersection">intersection</span></em>. Each element in the result must be <strong>unique</strong> and you may return the result in <strong>any order</strong>.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[English Version](/solution/0800-0899/0840.Magic%20Squares%20In%20Grid/README_EN.md)
|
||||
|
||||
<!-- tags:数组,数学,矩阵 -->
|
||||
<!-- tags:数组,哈希表,数学,矩阵 -->
|
||||
|
||||
## 题目描述
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[中文文档](/solution/0800-0899/0840.Magic%20Squares%20In%20Grid/README.md)
|
||||
|
||||
<!-- tags:Array,Math,Matrix -->
|
||||
<!-- tags:Array,Hash Table,Math,Matrix -->
|
||||
|
||||
## Description
|
||||
|
||||
|
|
|
|||
|
|
@ -54,6 +54,34 @@ Employees table:
|
|||
<strong>Explanation:</strong> Hercy has 2 people report directly to him, Alice and Bob. Their average age is (41+36)/2 = 38.5, which is 39 after rounding it to the nearest integer.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong>
|
||||
Employees table:
|
||||
+-------------+---------+------------+-----+
|
||||
| employee_id | name | reports_to | age |
|
||||
|-------------|---------|------------|-----|
|
||||
| 1 | Michael | null | 45 |
|
||||
| 2 | Alice | 1 | 38 |
|
||||
| 3 | Bob | 1 | 42 |
|
||||
| 4 | Charlie | 2 | 34 |
|
||||
| 5 | David | 2 | 40 |
|
||||
| 6 | Eve | 3 | 37 |
|
||||
| 7 | Frank | null | 50 |
|
||||
| 8 | Grace | null | 48 |
|
||||
+-------------+---------+------------+-----+
|
||||
<strong>Output:</strong>
|
||||
+-------------+---------+---------------+-------------+
|
||||
| employee_id | name | reports_count | average_age |
|
||||
| ----------- | ------- | ------------- | ----------- |
|
||||
| 1 | Michael | 2 | 40 |
|
||||
| 2 | Alice | 2 | 37 |
|
||||
| 3 | Bob | 1 | 37 |
|
||||
+-------------+---------+---------------+-------------+
|
||||
|
||||
</pre>
|
||||
|
||||
## Solutions
|
||||
|
||||
### Solution 1: Self-Join + Grouping
|
||||
|
|
|
|||
|
|
@ -48,7 +48,7 @@
|
|||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>evnet1.length == event2.length == 2.</code></li>
|
||||
<li>event1<code>.length == </code>event2<code>.length == 2.</code></li>
|
||||
<li><code>event1[i].length == event2[i].length == 5</code></li>
|
||||
<li><code>startTime<sub>1</sub> <= endTime<sub>1</sub></code></li>
|
||||
<li><code>startTime<sub>2</sub> <= endTime<sub>2</sub></code></li>
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[English Version](/solution/2600-2699/2617.Minimum%20Number%20of%20Visited%20Cells%20in%20a%20Grid/README_EN.md)
|
||||
|
||||
<!-- tags:栈,并查集,树状数组,线段树,数组,二分查找,动态规划 -->
|
||||
<!-- tags:栈,广度优先搜索,并查集,数组,动态规划,矩阵,单调栈,堆（优先队列） -->
|
||||
|
||||
## 题目描述
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[中文文档](/solution/2600-2699/2617.Minimum%20Number%20of%20Visited%20Cells%20in%20a%20Grid/README.md)
|
||||
|
||||
<!-- tags:Stack,Union Find,Binary Indexed Tree,Segment Tree,Array,Binary Search,Dynamic Programming -->
|
||||
<!-- tags:Stack,Breadth-First Search,Union Find,Array,Dynamic Programming,Matrix,Monotonic Stack,Heap (Priority Queue) -->
|
||||
|
||||
## Description
|
||||
|
||||
|
|
|
|||
|
|
@ -6,35 +6,42 @@
|
|||
|
||||
## Description
|
||||
|
||||
<p>You have <code>n</code> processors each having <code>4</code> cores and <code>n * 4</code> tasks that need to be executed such that each core should perform only <strong>one</strong> task.</p>
|
||||
<p>You have a certain number of processors, each having 4 cores. The number of tasks to be executed is four times the number of processors. Each task must be assigned to a unique core, and each core can only be used once.</p>
|
||||
|
||||
<p>Given a <strong>0-indexed</strong> integer array <code>processorTime</code> representing the time at which each processor becomes available for the first time and a <strong>0-indexed </strong>integer array <code>tasks</code> representing the time it takes to execute each task, return <em>the <strong>minimum</strong> time when all of the tasks have been executed by the processors.</em></p>
|
||||
|
||||
<p><strong>Note: </strong>Each core executes the task independently of the others.</p>
|
||||
<p>You are given an array <code>processorTime</code> representing the time each processor becomes available and an array <code>tasks</code> representing how long each task takes to complete. Return the <em>minimum</em> time needed to complete all tasks.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> processorTime = [8,10], tasks = [2,2,3,1,8,7,4,5]
|
||||
<strong>Output:</strong> 16
|
||||
<strong>Explanation:</strong>
|
||||
It's optimal to assign the tasks at indexes 4, 5, 6, 7 to the first processor which becomes available at time = 8, and the tasks at indexes 0, 1, 2, 3 to the second processor which becomes available at time = 10.
|
||||
Time taken by the first processor to finish execution of all tasks = max(8 + 8, 8 + 7, 8 + 4, 8 + 5) = 16.
|
||||
Time taken by the second processor to finish execution of all tasks = max(10 + 2, 10 + 2, 10 + 3, 10 + 1) = 13.
|
||||
Hence, it can be shown that the minimum time taken to execute all the tasks is 16.</pre>
|
||||
<div class="example-block">
|
||||
<p><strong>Input:</strong> <span class="example-io">processorTime = [8,10], tasks = [2,2,3,1,8,7,4,5]</span></p>
|
||||
|
||||
<p><strong>Output:</strong> <span class="example-io">16</span></p>
|
||||
|
||||
<p><strong>Explanation:</strong></p>
|
||||
|
||||
<p>Assign the tasks at indices 4, 5, 6, 7 to the first processor which becomes available at <code>time = 8</code>, and the tasks at indices 0, 1, 2, 3 to the second processor which becomes available at <code>time = 10</code>. </p>
|
||||
|
||||
<p>The time taken by the first processor to finish the execution of all tasks is <code>max(8 + 8, 8 + 7, 8 + 4, 8 + 5) = 16</code>.</p>
|
||||
|
||||
<p>The time taken by the second processor to finish the execution of all tasks is <code>max(10 + 2, 10 + 2, 10 + 3, 10 + 1) = 13</code>.</p>
|
||||
</div>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> processorTime = [10,20], tasks = [2,3,1,2,5,8,4,3]
|
||||
<strong>Output:</strong> 23
|
||||
<strong>Explanation:</strong>
|
||||
It's optimal to assign the tasks at indexes 1, 4, 5, 6 to the first processor which becomes available at time = 10, and the tasks at indexes 0, 2, 3, 7 to the second processor which becomes available at time = 20.
|
||||
Time taken by the first processor to finish execution of all tasks = max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18.
|
||||
Time taken by the second processor to finish execution of all tasks = max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23.
|
||||
Hence, it can be shown that the minimum time taken to execute all the tasks is 23.
|
||||
</pre>
|
||||
<div class="example-block">
|
||||
<p><strong>Input:</strong> <span class="example-io">processorTime = [10,20], tasks = [2,3,1,2,5,8,4,3]</span></p>
|
||||
|
||||
<p><strong>Output:</strong> <span class="example-io">23</span></p>
|
||||
|
||||
<p><strong>Explanation:</strong></p>
|
||||
|
||||
<p>Assign the tasks at indices 1, 4, 5, 6 to the first processor and the others to the second processor.</p>
|
||||
|
||||
<p>The time taken by the first processor to finish the execution of all tasks is <code>max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18</code>.</p>
|
||||
|
||||
<p>The time taken by the second processor to finish the execution of all tasks is <code>max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23</code>.</p>
|
||||
</div>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
|
|
|||
|
|
@ -8,9 +8,7 @@
|
|||
|
||||
<!-- 这里写题目描述 -->
|
||||
|
||||
<p>给你一个整数数组 <code>nums</code> ，其中 <code>nums[i]</code> 要么是一个正整数，要么是 <code>-1</code> 。</p>
|
||||
|
||||
<p>我们需要为每个 <code>-1</code> 找到相应的正整数，我们称之为最后访问的整数。</p>
|
||||
<p>给你一个整数数组 <code>nums</code> ，其中 <code>nums[i]</code> 要么是一个正整数，要么是 <code>-1</code> 。我们需要为每个 <code>-1</code> 找到相应的正整数，我们称之为最后访问的整数。</p>
|
||||
|
||||
<p>为了达到这个目标，定义两个空数组：<code>seen</code> 和 <code>ans</code>。</p>
|
||||
|
||||
|
|
|
|||
|
|
@ -6,9 +6,7 @@
|
|||
|
||||
## Description
|
||||
|
||||
<p>Given an integer array <code>nums</code> where <code>nums[i]</code> is either a positive integer or <code>-1</code>.</p>
|
||||
|
||||
<p>We need to find for each <code>-1</code> the respective positive integer, which we call the last visited integer.</p>
|
||||
<p>Given an integer array <code>nums</code> where <code>nums[i]</code> is either a positive integer or <code>-1</code>. We need to find for each <code>-1</code> the respective positive integer, which we call the last visited integer.</p>
|
||||
|
||||
<p>To achieve this goal, let's define two empty arrays: <code>seen</code> and <code>ans</code>.</p>
|
||||
|
||||
|
|
@ -24,27 +22,19 @@
|
|||
</li>
|
||||
</ul>
|
||||
|
||||
<p>Return <em>the array </em><code>ans</code>.</p>
|
||||
<p>Return the array<em> </em><code>ans</code>.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<div class="example-block" style="
|
||||
border-color: var(--border-tertiary);
|
||||
border-left-width: 2px;
|
||||
color: var(--text-secondary);
|
||||
font-size: .875rem;
|
||||
line-height: 1.25rem;
|
||||
margin-bottom: 1rem;
|
||||
margin-top: 1rem;
|
||||
overflow: visible;
|
||||
padding-left: 1rem;
|
||||
">
|
||||
<p><strong>Input:</strong> <code>nums = [1,2,-1,-1,-1]</code></p>
|
||||
<div class="example-block">
|
||||
<p><strong>Input:</strong> <span class="example-io">nums = [1,2,-1,-1,-1]</span></p>
|
||||
|
||||
<p><strong>Output:</strong> <code>[2,1,-1]</code></p>
|
||||
<p><strong>Output:</strong> <span class="example-io">[2,1,-1]</span></p>
|
||||
|
||||
<p><strong>Explanation:</strong> Start with <code>seen = []</code> and <code>ans = []</code>.</p>
|
||||
<p><strong>Explanation:</strong></p>
|
||||
|
||||
<p>Start with <code>seen = []</code> and <code>ans = []</code>.</p>
|
||||
|
||||
<ol>
|
||||
<li>Process <code>nums[0]</code>: The first element in nums is <code>1</code>. We prepend it to the front of <code>seen</code>. Now, <code>seen == [1]</code>.</li>
|
||||
|
|
@ -57,22 +47,14 @@
|
|||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<div class="example-block" style="
|
||||
border-color: var(--border-tertiary);
|
||||
border-left-width: 2px;
|
||||
color: var(--text-secondary);
|
||||
font-size: .875rem;
|
||||
line-height: 1.25rem;
|
||||
margin-bottom: 1rem;
|
||||
margin-top: 1rem;
|
||||
overflow: visible;
|
||||
padding-left: 1rem;
|
||||
">
|
||||
<p><strong>Input:</strong> <code>nums = [1,-1,2,-1,-1]</code></p>
|
||||
<div class="example-block">
|
||||
<p><strong>Input:</strong> <span class="example-io">nums = [1,-1,2,-1,-1]</span></p>
|
||||
|
||||
<p><strong>Output:</strong> <code>[1,2,1]</code></p>
|
||||
<p><strong>Output:</strong><span class="example-io"> [1,2,1]</span></p>
|
||||
|
||||
<p><strong>Explanation:</strong> Start with <code>seen = []</code> and <code>ans = []</code>.</p>
|
||||
<p><strong>Explanation:</strong></p>
|
||||
|
||||
<p>Start with <code>seen = []</code> and <code>ans = []</code>.</p>
|
||||
|
||||
<ol>
|
||||
<li>Process <code>nums[0]</code>: The first element in nums is <code>1</code>. We prepend it to the front of <code>seen</code>. Now, <code>seen == [1]</code>.</li>
|
||||
|
|
|
|||
|
|
@ -6,78 +6,56 @@
|
|||
|
||||
## Description
|
||||
|
||||
<p>There is a grid with <code>n + 2</code> <strong>horizontal</strong> bars and <code>m + 2</code> <strong>vertical</strong> bars, and initially containing <code>1 x 1</code> unit cells.</p>
|
||||
<p>You are given the two integers, <code>n</code> and <code>m</code> and two integer arrays, <code>hBars</code> and <code>vBars</code>. The grid has <code>n + 2</code> horizontal and <code>m + 2</code> vertical bars, creating 1 x 1 unit cells. The bars are indexed starting from <code>1</code>.</p>
|
||||
|
||||
<p>The bars are <strong>1-indexed</strong>.</p>
|
||||
<p>You can <strong>remove</strong> some of the bars in <code>hBars</code> from horizontal bars and some of the bars in <code>vBars</code> from vertical bars. Note that other bars are fixed and cannot be removed.</p>
|
||||
|
||||
<p>You are given the two integers, <code>n</code> and <code>m</code>.</p>
|
||||
|
||||
<p>You are also given two integer arrays: <code>hBars</code> and <code>vBars</code>.</p>
|
||||
|
||||
<ul>
|
||||
<li><code>hBars</code> contains <strong>distinct</strong> horizontal bars in the range <code>[2, n + 1]</code>.</li>
|
||||
<li><code>vBars</code> contains <strong>distinct</strong> vertical bars in the range <code>[2, m + 1]</code>.</li>
|
||||
</ul>
|
||||
|
||||
<p>You are allowed to <strong>remove</strong> bars that satisfy any of the following conditions:</p>
|
||||
|
||||
<ul>
|
||||
<li>If it is a horizontal bar, it must correspond to a value in <code>hBars</code>.</li>
|
||||
<li>If it is a vertical bar, it must correspond to a value in <code>vBars</code>.</li>
|
||||
</ul>
|
||||
|
||||
<p>Return <em>an integer denoting the <strong>maximum</strong> area of a <strong>square-shaped</strong> hole in the grid after removing some bars (<strong>possibly none</strong>).</em></p>
|
||||
<p>Return an integer denoting the <strong>maximum area</strong> of a <em>square-shaped</em> hole in the grid, after removing some bars (possibly none).</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2900-2999/2943.Maximize%20Area%20of%20Square%20Hole%20in%20Grid/images/screenshot-from-2023-11-05-22-40-25.png" style="width: 411px; height: 220px;" /></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 2, m = 1, hBars = [2,3], vBars = [2]
|
||||
<strong>Output:</strong> 4
|
||||
<strong>Explanation:</strong> The left image shows the initial grid formed by the bars.
|
||||
The horizontal bars are in the range [1,4], and the vertical bars are in the range [1,3].
|
||||
It is allowed to remove horizontal bars [2,3] and the vertical bar [2].
|
||||
One way to get the maximum square-shaped hole is by removing horizontal bar 2 and vertical bar 2.
|
||||
The resulting grid is shown on the right.
|
||||
The hole has an area of 4.
|
||||
It can be shown that it is not possible to get a square hole with an area more than 4.
|
||||
Hence, the answer is 4.
|
||||
</pre>
|
||||
<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
|
||||
<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">n = 2, m = 1, hBars = [2,3], vBars = [2]</span></p>
|
||||
|
||||
<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">4</span></p>
|
||||
|
||||
<p><strong>Explanation:</strong></p>
|
||||
|
||||
<p>The left image shows the initial grid formed by the bars. The horizontal bars are <code>[1,2,3,4]</code>, and the vertical bars are <code>[1,2,3]</code>.</p>
|
||||
|
||||
<p>One way to get the maximum square-shaped hole is by removing horizontal bar 2 and vertical bar 2.</p>
|
||||
</div>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2900-2999/2943.Maximize%20Area%20of%20Square%20Hole%20in%20Grid/images/screenshot-from-2023-11-04-17-01-02.png" style="width: 368px; height: 145px;" /></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 1, m = 1, hBars = [2], vBars = [2]
|
||||
<strong>Output:</strong> 4
|
||||
<strong>Explanation:</strong> The left image shows the initial grid formed by the bars.
|
||||
The horizontal bars are in the range [1,3], and the vertical bars are in the range [1,3].
|
||||
It is allowed to remove the horizontal bar [2] and the vertical bar [2].
|
||||
To get the maximum square-shaped hole, we remove horizontal bar 2 and vertical bar 2.
|
||||
The resulting grid is shown on the right.
|
||||
The hole has an area of 4.
|
||||
Hence, the answer is 4, and it is the maximum possible.
|
||||
</pre>
|
||||
<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
|
||||
<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">n = 1, m = 1, hBars = [2], vBars = [2]</span></p>
|
||||
|
||||
<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">4</span></p>
|
||||
|
||||
<p><strong>Explanation:</strong></p>
|
||||
|
||||
<p>To get the maximum square-shaped hole, we remove horizontal bar 2 and vertical bar 2.</p>
|
||||
</div>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2900-2999/2943.Maximize%20Area%20of%20Square%20Hole%20in%20Grid/images/screenshot-from-2023-11-05-22-33-35.png" style="width: 648px; height: 218px;" /></p>
|
||||
<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2900-2999/2943.Maximize%20Area%20of%20Square%20Hole%20in%20Grid/images/unsaved-image-2.png" style="width: 648px; height: 218px;" /></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 2, m = 3, hBars = [2,3], vBars = [2,3,4]
|
||||
<strong>Output:</strong> 9
|
||||
<strong>Explanation:</strong> The left image shows the initial grid formed by the bars.
|
||||
The horizontal bars are in the range [1,4], and the vertical bars are in the range [1,5].
|
||||
It is allowed to remove horizontal bars [2,3] and vertical bars [2,3,4].
|
||||
One way to get the maximum square-shaped hole is by removing horizontal bars 2 and 3, and vertical bars 3 and 4.
|
||||
The resulting grid is shown on the right.
|
||||
The hole has an area of 9.
|
||||
It can be shown that it is not possible to get a square hole with an area more than 9.
|
||||
Hence, the answer is 9.
|
||||
</pre>
|
||||
<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
|
||||
<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">n = 2, m = 3, hBars = [2,3], vBars = [2,4]</span></p>
|
||||
|
||||
<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">4</span></p>
|
||||
|
||||
<p><strong>Explanation:</strong></p>
|
||||
|
||||
<p><span style="color: var(--text-secondary); font-size: 0.875rem;">One way to get the maximum square-shaped hole is by removing horizontal bar 3, and vertical bar 4.</span></p>
|
||||
</div>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
|
|
|||
|
|
@ -8,18 +8,54 @@
|
|||
|
||||
<!-- 这里写题目描述 -->
|
||||
|
||||
<p>给你一个整数 <code>k</code> 和一个整数 <code>x</code> 。</p>
|
||||
<p>给你一个整数 <code>k</code> 和一个整数 <code>x</code> 。整数 <code>num</code> 的价值是由它的二进制表示中，从最低有效位开始，<code>x</code>，<code>2x</code>，<code>3x</code>，以此类推，这些位置上 <strong>设置位</strong> 的数目来计算。下面的表格包含了如何计算价值的例子。</p>
|
||||
|
||||
<p>令 <code>s</code> 为整数 <code>num</code> 的下标从 <strong>1</strong> 开始的二进制表示。我们说一个整数 <code>num</code> 的 <strong>价值</strong> 是满足 <code>i % x == 0</code> 且 <code><font face="monospace">s[i]</font></code> 是 <strong>设置位</strong> 的 <code>i</code> 的数目。</p>
|
||||
<table border="1">
|
||||
<tbody>
|
||||
<tr>
|
||||
<th>x</th>
|
||||
<th>num</th>
|
||||
<th>Binary Representation</th>
|
||||
<th>Price</th>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>13</td>
|
||||
<td><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><strong><u>1</u></strong><strong><u>1</u></strong><u>0</u><strong><u>1</u></strong></td>
|
||||
<td>3</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>13</td>
|
||||
<td>0<u>0</u>0<u>0</u>0<strong><u>1</u></strong>1<u>0</u>1</td>
|
||||
<td>1</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>233</td>
|
||||
<td>0<strong><u>1</u></strong>1<strong><u>1</u></strong>0<strong><u>1</u></strong>0<u>0</u>1</td>
|
||||
<td>3</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>3</td>
|
||||
<td>13</td>
|
||||
<td><u>0</u>00<u>0</u>01<strong><u>1</u></strong>01</td>
|
||||
<td>1</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>3</td>
|
||||
<td>362</td>
|
||||
<td><strong><u>1</u></strong>01<strong><u>1</u></strong>01<u>0</u>10</td>
|
||||
<td>2</td>
|
||||
</tr>
|
||||
</tbody>
|
||||
</table>
|
||||
|
||||
<p>请你返回<strong> 最大</strong> 整数<em> </em><code>num</code> ，满足从 <code>1</code> 到 <code>num</code> 的所有整数的 <strong>价值</strong> 和小于等于 <code>k</code> 。</p>
|
||||
<p> </p>
|
||||
|
||||
<p><b>注意：</b></p>
|
||||
<p><code>num</code> 的 <strong>累加价值</strong> 是从 <code>1</code> 到 <code>num</code> 的数字的 <strong>总</strong> 价值。如果 <code>num</code> 的累加价值小于或等于 <code>k</code> 则被认为是 <strong>廉价</strong> 的。</p>
|
||||
|
||||
<ul>
|
||||
<li>一个整数二进制表示下 <strong>设置位</strong> 是值为 <code>1</code> 的数位。</li>
|
||||
<li>一个整数的二进制表示下标从右到左编号，比方说如果 <code>s == 11100</code> ，那么 <code>s[4] == 1</code> 且 <code>s[2] == 0</code> 。</li>
|
||||
</ul>
|
||||
<p>请你返回<strong> 最大</strong> 的廉价数字。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
|
|
@ -28,24 +64,159 @@
|
|||
<pre>
|
||||
<b>输入：</b>k = 9, x = 1
|
||||
<b>输出：</b>6
|
||||
<b>解释：</b>数字 1 ，2 ，3 ，4 ，5 和 6 二进制表示分别为 "1" ，"10" ，"11" ，"100" ，"101" 和 "110" 。
|
||||
由于 x 等于 1 ，每个数字的价值分别为所有设置位的数目。
|
||||
这些数字的所有设置位数目总数是 9 ，所以前 6 个数字的价值和为 9 。
|
||||
所以答案为 6 。</pre>
|
||||
<b>解释：</b>由下表所示，6 是最大的廉价数字。
|
||||
</pre>
|
||||
|
||||
<table border="1">
|
||||
<tbody>
|
||||
<tr>
|
||||
<th>x</th>
|
||||
<th>num</th>
|
||||
<th>Binary Representation</th>
|
||||
<th>Price</th>
|
||||
<th>Accumulated Price</th>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>1</td>
|
||||
<td><u>0</u><u>0</u><strong><u>1</u></strong></td>
|
||||
<td>1</td>
|
||||
<td>1</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>2</td>
|
||||
<td><u>0</u><strong><u>1</u></strong><u>0</u></td>
|
||||
<td>1</td>
|
||||
<td>2</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>3</td>
|
||||
<td><u>0</u><strong><u>1</u></strong><strong><u>1</u></strong></td>
|
||||
<td>2</td>
|
||||
<td>4</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>4</td>
|
||||
<td><strong><u>1</u></strong><u>0</u><u>0</u></td>
|
||||
<td>1</td>
|
||||
<td>5</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>5</td>
|
||||
<td><strong><u>1</u></strong><u>0</u><strong><u>1</u></strong></td>
|
||||
<td>2</td>
|
||||
<td>7</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>6</td>
|
||||
<td><strong><u>1</u></strong><strong><u>1</u></strong><u>0</u></td>
|
||||
<td>2</td>
|
||||
<td>9</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>7</td>
|
||||
<td><strong><u>1</u></strong><strong><u>1</u></strong><strong><u>1</u></strong></td>
|
||||
<td>3</td>
|
||||
<td>12</td>
|
||||
</tr>
|
||||
</tbody>
|
||||
</table>
|
||||
|
||||
<p><strong class="example">示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>k = 7, x = 2
|
||||
<b>输出：</b>9
|
||||
<b>解释：</b>由于 x 等于 2 ，我们检查每个数字的偶数位。
|
||||
2 和 3 在二进制表示下的第二个数位为设置位，所以它们的价值和为 2 。
|
||||
6 和 7 在二进制表示下的第二个数位为设置位，所以它们的价值和为 2 。
|
||||
8 和 9 在二进制表示下的第四个数位为设置位但第二个数位不是设置位，所以它们的价值和为 2 。
|
||||
数字 1 ，4 和 5 在二进制下偶数位都不是设置位，所以它们的价值和为 0 。
|
||||
10 在二进制表示下的第二个数位和第四个数位都是设置位，所以它的价值为 2 。
|
||||
前 9 个数字的价值和为 6 。
|
||||
前 10 个数字的价值和为 8，超过了 k = 7 ，所以答案为 9 。</pre>
|
||||
<b>解释：</b>由下表所示，9 是最大的廉价数字。
|
||||
</pre>
|
||||
|
||||
<table border="1">
|
||||
<tbody>
|
||||
<tr>
|
||||
<th>x</th>
|
||||
<th>num</th>
|
||||
<th>Binary Representation</th>
|
||||
<th>Price</th>
|
||||
<th>Accumulated Price</th>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>1</td>
|
||||
<td><u>0</u>0<u>0</u>1</td>
|
||||
<td>0</td>
|
||||
<td>0</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>2</td>
|
||||
<td><u>0</u>0<strong><u>1</u></strong>0</td>
|
||||
<td>1</td>
|
||||
<td>1</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>3</td>
|
||||
<td><u>0</u>0<strong><u>1</u></strong>1</td>
|
||||
<td>1</td>
|
||||
<td>2</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>4</td>
|
||||
<td><u>0</u>1<u>0</u>0</td>
|
||||
<td>0</td>
|
||||
<td>2</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>5</td>
|
||||
<td><u>0</u>1<u>0</u>1</td>
|
||||
<td>0</td>
|
||||
<td>2</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>6</td>
|
||||
<td><u>0</u>1<strong><u>1</u></strong>0</td>
|
||||
<td>1</td>
|
||||
<td>3</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>7</td>
|
||||
<td><u>0</u>1<strong><u>1</u></strong>1</td>
|
||||
<td>1</td>
|
||||
<td>4</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>8</td>
|
||||
<td><strong><u>1</u></strong>0<u>0</u>0</td>
|
||||
<td>1</td>
|
||||
<td>5</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>9</td>
|
||||
<td><strong><u>1</u></strong>0<u>0</u>1</td>
|
||||
<td>1</td>
|
||||
<td>6</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>10</td>
|
||||
<td><strong><u>1</u></strong>0<strong><u>1</u></strong>0</td>
|
||||
<td>2</td>
|
||||
<td>8</td>
|
||||
</tr>
|
||||
</tbody>
|
||||
</table>
|
||||
|
||||
<p> </p>
|
||||
|
||||
|
|
|
|||
|
|
@ -6,43 +6,220 @@
|
|||
|
||||
## Description
|
||||
|
||||
<p>You are given an integer <code>k</code> and an integer <code>x</code>.</p>
|
||||
<p>You are given an integer <code>k</code> and an integer <code>x</code>. The price of a number <code>num</code> is calculated by the count of <span data-keyword="set-bit">set bits</span> at positions <code>x</code>, <code>2x</code>, <code>3x</code>, etc., in its binary representation, starting from the least significant bit. The following table contains examples of how price is calculated.</p>
|
||||
|
||||
<p>Consider <code>s</code> is the <strong>1-indexed </strong>binary representation of an integer <code>num</code>. The <strong>price</strong> of a number <code>num</code> is the number of <code>i</code>'s such that <code>i % x == 0</code> and <code><font face="monospace">s[i]</font></code> is a <strong>set bit</strong>.</p>
|
||||
<table border="1">
|
||||
<tbody>
|
||||
<tr>
|
||||
<th>x</th>
|
||||
<th>num</th>
|
||||
<th>Binary Representation</th>
|
||||
<th>Price</th>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>13</td>
|
||||
<td><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><strong><u>1</u></strong><strong><u>1</u></strong><u>0</u><strong><u>1</u></strong></td>
|
||||
<td>3</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>13</td>
|
||||
<td>0<u>0</u>0<u>0</u>0<strong><u>1</u></strong>1<u>0</u>1</td>
|
||||
<td>1</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>233</td>
|
||||
<td>0<strong><u>1</u></strong>1<strong><u>1</u></strong>0<strong><u>1</u></strong>0<u>0</u>1</td>
|
||||
<td>3</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>3</td>
|
||||
<td>13</td>
|
||||
<td><u>0</u>00<u>0</u>01<strong><u>1</u></strong>01</td>
|
||||
<td>1</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>3</td>
|
||||
<td>362</td>
|
||||
<td><strong><u>1</u></strong>01<strong><u>1</u></strong>01<u>0</u>10</td>
|
||||
<td>2</td>
|
||||
</tr>
|
||||
</tbody>
|
||||
</table>
|
||||
|
||||
<p>Return <em>the <b>greatest</b> integer </em><code>num</code><em> such that the sum of <strong>prices</strong> of all numbers from </em><code>1</code><em> to </em><code>num</code><em> is less than or equal to </em><code>k</code><em>.</em></p>
|
||||
<p>The <strong>accumulated price</strong> of <code>num</code> is the <b>total</b> price of numbers from <code>1</code> to <code>num</code>. <code>num</code> is considered <strong>cheap</strong> if its accumulated price is less than or equal to <code>k</code>.</p>
|
||||
|
||||
<p><strong>Note</strong>:</p>
|
||||
|
||||
<ul>
|
||||
<li>In the binary representation of a number <strong>set bit</strong> is a bit of value <code>1</code>.</li>
|
||||
<li>The binary representation of a number will be indexed from right to left. For example, if <code>s == 11100</code>, <code>s[4] == 1</code> and <code>s[2] == 0</code>.</li>
|
||||
</ul>
|
||||
<p>Return the <b>greatest</b> cheap number.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> k = 9, x = 1
|
||||
<strong>Output:</strong> 6
|
||||
<strong>Explanation:</strong> The numbers 1, 2, 3, 4, 5, and 6 can be written in binary representation as "1", "10", "11", "100", "101", and "110" respectively.
|
||||
Since x is equal to 1, the price of each number is the number of its set bits.
|
||||
The number of set bits in these numbers is 9. So the sum of the prices of the first 6 numbers is 9.
|
||||
So the answer is 6.</pre>
|
||||
<div class="example-block">
|
||||
<p><strong>Input:</strong> <span class="example-io">k = 9, x = 1</span></p>
|
||||
|
||||
<p><strong>Output:</strong> <span class="example-io">6</span></p>
|
||||
|
||||
<p><strong>Explanation:</strong></p>
|
||||
|
||||
<p>As shown in the table below, <code>6</code> is the greatest cheap number.</p>
|
||||
|
||||
<table border="1">
|
||||
<tbody>
|
||||
<tr>
|
||||
<th>x</th>
|
||||
<th>num</th>
|
||||
<th>Binary Representation</th>
|
||||
<th>Price</th>
|
||||
<th>Accumulated Price</th>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>1</td>
|
||||
<td><u>0</u><u>0</u><strong><u>1</u></strong></td>
|
||||
<td>1</td>
|
||||
<td>1</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>2</td>
|
||||
<td><u>0</u><strong><u>1</u></strong><u>0</u></td>
|
||||
<td>1</td>
|
||||
<td>2</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>3</td>
|
||||
<td><u>0</u><strong><u>1</u></strong><strong><u>1</u></strong></td>
|
||||
<td>2</td>
|
||||
<td>4</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>4</td>
|
||||
<td><strong><u>1</u></strong><u>0</u><u>0</u></td>
|
||||
<td>1</td>
|
||||
<td>5</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>5</td>
|
||||
<td><strong><u>1</u></strong><u>0</u><strong><u>1</u></strong></td>
|
||||
<td>2</td>
|
||||
<td>7</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>6</td>
|
||||
<td><strong><u>1</u></strong><strong><u>1</u></strong><u>0</u></td>
|
||||
<td>2</td>
|
||||
<td>9</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>1</td>
|
||||
<td>7</td>
|
||||
<td><strong><u>1</u></strong><strong><u>1</u></strong><strong><u>1</u></strong></td>
|
||||
<td>3</td>
|
||||
<td>12</td>
|
||||
</tr>
|
||||
</tbody>
|
||||
</table>
|
||||
</div>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> k = 7, x = 2
|
||||
<strong>Output:</strong> 9
|
||||
<strong>Explanation:</strong> Since x is equal to 2, we should just check even<sup>th</sup> bits.
|
||||
The second bit of binary representation of numbers 2 and 3 is a set bit. So the sum of their prices is 2.
|
||||
The second bit of binary representation of numbers 6 and 7 is a set bit. So the sum of their prices is 2.
|
||||
The fourth bit of binary representation of numbers 8 and 9 is a set bit but their second bit is not. So the sum of their prices is 2.
|
||||
Numbers 1, 4, and 5 don't have set bits in their even<sup>th</sup> bits in their binary representation. So the sum of their prices is 0.
|
||||
The second and the fourth bit of the binary representation of the number 10 are a set bit. So its price is 2.
|
||||
The sum of the prices of the first 9 numbers is 6.
|
||||
Because the sum of the prices of the first 10 numbers is 8, the answer is 9.</pre>
|
||||
<div class="example-block">
|
||||
<p><strong>Input:</strong> <span class="example-io">k = 7, x = 2</span></p>
|
||||
|
||||
<p><strong>Output:</strong> <span class="example-io">9</span></p>
|
||||
|
||||
<p><strong>Explanation:</strong></p>
|
||||
|
||||
<p>As shown in the table below, <code>9</code> is the greatest cheap number.</p>
|
||||
|
||||
<table border="1">
|
||||
<tbody>
|
||||
<tr>
|
||||
<th>x</th>
|
||||
<th>num</th>
|
||||
<th>Binary Representation</th>
|
||||
<th>Price</th>
|
||||
<th>Accumulated Price</th>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>1</td>
|
||||
<td><u>0</u>0<u>0</u>1</td>
|
||||
<td>0</td>
|
||||
<td>0</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>2</td>
|
||||
<td><u>0</u>0<strong><u>1</u></strong>0</td>
|
||||
<td>1</td>
|
||||
<td>1</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>3</td>
|
||||
<td><u>0</u>0<strong><u>1</u></strong>1</td>
|
||||
<td>1</td>
|
||||
<td>2</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>4</td>
|
||||
<td><u>0</u>1<u>0</u>0</td>
|
||||
<td>0</td>
|
||||
<td>2</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>5</td>
|
||||
<td><u>0</u>1<u>0</u>1</td>
|
||||
<td>0</td>
|
||||
<td>2</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>6</td>
|
||||
<td><u>0</u>1<strong><u>1</u></strong>0</td>
|
||||
<td>1</td>
|
||||
<td>3</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>7</td>
|
||||
<td><u>0</u>1<strong><u>1</u></strong>1</td>
|
||||
<td>1</td>
|
||||
<td>4</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>8</td>
|
||||
<td><strong><u>1</u></strong>0<u>0</u>0</td>
|
||||
<td>1</td>
|
||||
<td>5</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>9</td>
|
||||
<td><strong><u>1</u></strong>0<u>0</u>1</td>
|
||||
<td>1</td>
|
||||
<td>6</td>
|
||||
</tr>
|
||||
<tr>
|
||||
<td>2</td>
|
||||
<td>10</td>
|
||||
<td><strong><u>1</u></strong>0<strong><u>1</u></strong>0</td>
|
||||
<td>2</td>
|
||||
<td>8</td>
|
||||
</tr>
|
||||
</tbody>
|
||||
</table>
|
||||
</div>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
|
|
|||
|
|
@ -1,4 +1,4 @@
|
|||
# [3050. Pizza Toppings Cost Analysis](https://leetcode.cn/problems/pizza-toppings-cost-analysis)
|
||||
# [3050. 披萨配料成本分析](https://leetcode.cn/problems/pizza-toppings-cost-analysis)
|
||||
|
||||
[English Version](/solution/3000-3099/3050.Pizza%20Toppings%20Cost%20Analysis/README_EN.md)
|
||||
|
||||
|
|
|
|||
|
|
@ -1,4 +1,4 @@
|
|||
# [3051. Find Candidates for Data Scientist Position](https://leetcode.cn/problems/find-candidates-for-data-scientist-position)
|
||||
# [3051. 寻找数据科学家职位的候选人](https://leetcode.cn/problems/find-candidates-for-data-scientist-position)
|
||||
|
||||
[English Version](/solution/3000-3099/3051.Find%20Candidates%20for%20Data%20Scientist%20Position/README_EN.md)
|
||||
|
||||
|
|
|
|||
|
|
@ -1,4 +1,4 @@
|
|||
# [3052. Maximize Items](https://leetcode.cn/problems/maximize-items)
|
||||
# [3052. 最大化商品](https://leetcode.cn/problems/maximize-items)
|
||||
|
||||
[English Version](/solution/3000-3099/3052.Maximize%20Items/README_EN.md)
|
||||
|
||||
|
|
|
|||
|
|
@ -1,4 +1,4 @@
|
|||
# [3053. Classifying Triangles by Lengths](https://leetcode.cn/problems/classifying-triangles-by-lengths)
|
||||
# [3053. 根据长度分类三角形](https://leetcode.cn/problems/classifying-triangles-by-lengths)
|
||||
|
||||
[English Version](/solution/3000-3099/3053.Classifying%20Triangles%20by%20Lengths/README_EN.md)
|
||||
|
||||
|
|
|
|||
|
|
@ -1,4 +1,4 @@
|
|||
# [3054. Binary Tree Nodes](https://leetcode.cn/problems/binary-tree-nodes)
|
||||
# [3054. 二叉树节点](https://leetcode.cn/problems/binary-tree-nodes)
|
||||
|
||||
[English Version](/solution/3000-3099/3054.Binary%20Tree%20Nodes/README_EN.md)
|
||||
|
||||
|
|
|
|||
|
|
@ -1,4 +1,4 @@
|
|||
# [3055. Top Percentile Fraud](https://leetcode.cn/problems/top-percentile-fraud)
|
||||
# [3055. 最高欺诈百分位数](https://leetcode.cn/problems/top-percentile-fraud)
|
||||
|
||||
[English Version](/solution/3000-3099/3055.Top%20Percentile%20Fraud/README_EN.md)
|
||||
|
||||
|
|
|
|||
|
|
@ -1,4 +1,4 @@
|
|||
# [3056. Snaps Analysis](https://leetcode.cn/problems/snaps-analysis)
|
||||
# [3056. 快照分析](https://leetcode.cn/problems/snaps-analysis)
|
||||
|
||||
[English Version](/solution/3000-3099/3056.Snaps%20Analysis/README_EN.md)
|
||||
|
||||
|
|
|
|||
|
|
@ -1,4 +1,4 @@
|
|||
# [3057. Employees Project Allocation](https://leetcode.cn/problems/employees-project-allocation)
|
||||
# [3057. 员工项目分配](https://leetcode.cn/problems/employees-project-allocation)
|
||||
|
||||
[English Version](/solution/3000-3099/3057.Employees%20Project%20Allocation/README_EN.md)
|
||||
|
||||
|
|
|
|||
|
|
@ -1,4 +1,4 @@
|
|||
# [3058. Friends With No Mutual Friends](https://leetcode.cn/problems/friends-with-no-mutual-friends)
|
||||
# [3058. 没有共同朋友的朋友](https://leetcode.cn/problems/friends-with-no-mutual-friends)
|
||||
|
||||
[English Version](/solution/3000-3099/3058.Friends%20With%20No%20Mutual%20Friends/README_EN.md)
|
||||
|
||||
|
|
|
|||
|
|
@ -1,4 +1,4 @@
|
|||
# [3059. Find All Unique Email Domains](https://leetcode.cn/problems/find-all-unique-email-domains)
|
||||
# [3059. 找到所有不同的邮件域名](https://leetcode.cn/problems/find-all-unique-email-domains)
|
||||
|
||||
[English Version](/solution/3000-3099/3059.Find%20All%20Unique%20Email%20Domains/README_EN.md)
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[English Version](/solution/3000-3099/3087.Find%20Trending%20Hashtags/README_EN.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:数据库 -->
|
||||
|
||||
## 题目描述
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[中文文档](/solution/3000-3099/3087.Find%20Trending%20Hashtags/README.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:Database -->
|
||||
|
||||
## Description
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[English Version](/solution/3000-3099/3088.Make%20String%20Anti-palindrome/README_EN.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:贪心,字符串,排序 -->
|
||||
|
||||
## 题目描述
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[中文文档](/solution/3000-3099/3088.Make%20String%20Anti-palindrome/README.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:Greedy,String,Sorting -->
|
||||
|
||||
## Description
|
||||
|
||||
|
|
|
|||
|
|
@ -1,8 +1,8 @@
|
|||
# [3089. Find Bursty Behavior](https://leetcode.cn/problems/find-bursty-behavior)
|
||||
# [3089. 查找突发行为](https://leetcode.cn/problems/find-bursty-behavior)
|
||||
|
||||
[English Version](/solution/3000-3099/3089.Find%20Bursty%20Behavior/README_EN.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:数据库 -->
|
||||
|
||||
## 题目描述
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[中文文档](/solution/3000-3099/3089.Find%20Bursty%20Behavior/README.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:Database -->
|
||||
|
||||
## Description
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[English Version](/solution/3000-3099/3090.Maximum%20Length%20Substring%20With%20Two%20Occurrences/README_EN.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:哈希表,字符串,滑动窗口 -->
|
||||
|
||||
## 题目描述
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[中文文档](/solution/3000-3099/3090.Maximum%20Length%20Substring%20With%20Two%20Occurrences/README.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:Hash Table,String,Sliding Window -->
|
||||
|
||||
## Description
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[English Version](/solution/3000-3099/3091.Apply%20Operations%20to%20Make%20Sum%20of%20Array%20Greater%20Than%20or%20Equal%20to%20k/README_EN.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:贪心,数学,枚举 -->
|
||||
|
||||
## 题目描述
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[中文文档](/solution/3000-3099/3091.Apply%20Operations%20to%20Make%20Sum%20of%20Array%20Greater%20Than%20or%20Equal%20to%20k/README.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:Greedy,Math,Enumeration -->
|
||||
|
||||
## Description
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[English Version](/solution/3000-3099/3092.Most%20Frequent%20IDs/README_EN.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:数组,哈希表,有序集合,堆（优先队列） -->
|
||||
|
||||
## 题目描述
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[中文文档](/solution/3000-3099/3092.Most%20Frequent%20IDs/README.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:Array,Hash Table,Ordered Set,Heap (Priority Queue) -->
|
||||
|
||||
## Description
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[English Version](/solution/3000-3099/3093.Longest%20Common%20Suffix%20Queries/README_EN.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:字典树,数组,字符串 -->
|
||||
|
||||
## 题目描述
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
[中文文档](/solution/3000-3099/3093.Longest%20Common%20Suffix%20Queries/README.md)
|
||||
|
||||
<!-- tags: -->
|
||||
<!-- tags:Trie,Array,String -->
|
||||
|
||||
## Description
|
||||
|
||||
|
|
|
|||
|
|
@ -259,20 +259,20 @@
|
|||
| 2993 | [发生在周五的交易 I](/solution/2900-2999/2993.Friday%20Purchases%20I/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 2994 | [发生在周五的交易 II](/solution/2900-2999/2994.Friday%20Purchases%20II/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 2995 | [观众变主播](/solution/2900-2999/2995.Viewers%20Turned%20Streamers/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 3050 | [Pizza Toppings Cost Analysis](/solution/3000-3099/3050.Pizza%20Toppings%20Cost%20Analysis/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3051 | [Find Candidates for Data Scientist Position](/solution/3000-3099/3051.Find%20Candidates%20for%20Data%20Scientist%20Position/README.md) | `数据库` | 简单 | 🔒 |
|
||||
| 3052 | [Maximize Items](/solution/3000-3099/3052.Maximize%20Items/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 3053 | [Classifying Triangles by Lengths](/solution/3000-3099/3053.Classifying%20Triangles%20by%20Lengths/README.md) | `数据库` | 简单 | 🔒 |
|
||||
| 3054 | [Binary Tree Nodes](/solution/3000-3099/3054.Binary%20Tree%20Nodes/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3055 | [Top Percentile Fraud](/solution/3000-3099/3055.Top%20Percentile%20Fraud/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3056 | [Snaps Analysis](/solution/3000-3099/3056.Snaps%20Analysis/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3057 | [Employees Project Allocation](/solution/3000-3099/3057.Employees%20Project%20Allocation/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 3058 | [Friends With No Mutual Friends](/solution/3000-3099/3058.Friends%20With%20No%20Mutual%20Friends/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3059 | [Find All Unique Email Domains](/solution/3000-3099/3059.Find%20All%20Unique%20Email%20Domains/README.md) | `数据库` | 简单 | 🔒 |
|
||||
| 3050 | [披萨配料成本分析](/solution/3000-3099/3050.Pizza%20Toppings%20Cost%20Analysis/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3051 | [寻找数据科学家职位的候选人](/solution/3000-3099/3051.Find%20Candidates%20for%20Data%20Scientist%20Position/README.md) | `数据库` | 简单 | 🔒 |
|
||||
| 3052 | [最大化商品](/solution/3000-3099/3052.Maximize%20Items/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 3053 | [根据长度分类三角形](/solution/3000-3099/3053.Classifying%20Triangles%20by%20Lengths/README.md) | `数据库` | 简单 | 🔒 |
|
||||
| 3054 | [二叉树节点](/solution/3000-3099/3054.Binary%20Tree%20Nodes/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3055 | [最高欺诈百分位数](/solution/3000-3099/3055.Top%20Percentile%20Fraud/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3056 | [快照分析](/solution/3000-3099/3056.Snaps%20Analysis/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3057 | [员工项目分配](/solution/3000-3099/3057.Employees%20Project%20Allocation/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 3058 | [没有共同朋友的朋友](/solution/3000-3099/3058.Friends%20With%20No%20Mutual%20Friends/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3059 | [找到所有不同的邮件域名](/solution/3000-3099/3059.Find%20All%20Unique%20Email%20Domains/README.md) | `数据库` | 简单 | 🔒 |
|
||||
| 3060 | [时间范围内的用户活动](/solution/3000-3099/3060.User%20Activities%20within%20Time%20Bounds/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 3061 | [计算滞留雨水](/solution/3000-3099/3061.Calculate%20Trapping%20Rain%20Water/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 3087 | [查找热门话题标签](/solution/3000-3099/3087.Find%20Trending%20Hashtags/README.md) | | 中等 | 🔒 |
|
||||
| 3089 | [Find Bursty Behavior](/solution/3000-3099/3089.Find%20Bursty%20Behavior/README.md) | | 中等 | 🔒 |
|
||||
| 3087 | [查找热门话题标签](/solution/3000-3099/3087.Find%20Trending%20Hashtags/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3089 | [查找突发行为](/solution/3000-3099/3089.Find%20Bursty%20Behavior/README.md) | `数据库` | 中等 | 🔒 |
|
||||
|
||||
## 版权
|
||||
|
||||
|
|
|
|||
|
|
@ -269,8 +269,8 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
|
|||
| 3059 | [Find All Unique Email Domains](/solution/3000-3099/3059.Find%20All%20Unique%20Email%20Domains/README_EN.md) | `Database` | Easy | 🔒 |
|
||||
| 3060 | [User Activities within Time Bounds](/solution/3000-3099/3060.User%20Activities%20within%20Time%20Bounds/README_EN.md) | `Database` | Hard | 🔒 |
|
||||
| 3061 | [Calculate Trapping Rain Water](/solution/3000-3099/3061.Calculate%20Trapping%20Rain%20Water/README_EN.md) | `Database` | Hard | 🔒 |
|
||||
| 3087 | [Find Trending Hashtags](/solution/3000-3099/3087.Find%20Trending%20Hashtags/README_EN.md) | | Medium | 🔒 |
|
||||
| 3089 | [Find Bursty Behavior](/solution/3000-3099/3089.Find%20Bursty%20Behavior/README_EN.md) | | Medium | 🔒 |
|
||||
| 3087 | [Find Trending Hashtags](/solution/3000-3099/3087.Find%20Trending%20Hashtags/README_EN.md) | `Database` | Medium | 🔒 |
|
||||
| 3089 | [Find Bursty Behavior](/solution/3000-3099/3089.Find%20Bursty%20Behavior/README_EN.md) | `Database` | Medium | 🔒 |
|
||||
|
||||
## Copyright
|
||||
|
||||
|
|
|
|||
|
|
@ -331,7 +331,7 @@
|
|||
| 0318 | [最大单词长度乘积](/solution/0300-0399/0318.Maximum%20Product%20of%20Word%20Lengths/README.md) | `位运算`,`数组`,`字符串` | 中等 | |
|
||||
| 0319 | [灯泡开关](/solution/0300-0399/0319.Bulb%20Switcher/README.md) | `脑筋急转弯`,`数学` | 中等 | |
|
||||
| 0320 | [列举单词的全部缩写](/solution/0300-0399/0320.Generalized%20Abbreviation/README.md) | `位运算`,`字符串`,`回溯` | 中等 | 🔒 |
|
||||
| 0321 | [拼接最大数](/solution/0300-0399/0321.Create%20Maximum%20Number/README.md) | `栈`,`贪心`,`单调栈` | 困难 | |
|
||||
| 0321 | [拼接最大数](/solution/0300-0399/0321.Create%20Maximum%20Number/README.md) | `栈`,`贪心`,`数组`,`双指针`,`单调栈` | 困难 | |
|
||||
| 0322 | [零钱兑换](/solution/0300-0399/0322.Coin%20Change/README.md) | `广度优先搜索`,`数组`,`动态规划` | 中等 | |
|
||||
| 0323 | [无向图中连通分量的数目](/solution/0300-0399/0323.Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph/README.md) | `深度优先搜索`,`广度优先搜索`,`并查集`,`图` | 中等 | 🔒 |
|
||||
| 0324 | [摆动排序 II](/solution/0300-0399/0324.Wiggle%20Sort%20II/README.md) | `数组`,`分治`,`快速选择`,`排序` | 中等 | |
|
||||
|
|
@ -850,7 +850,7 @@
|
|||
| 0837 | [新 21 点](/solution/0800-0899/0837.New%2021%20Game/README.md) | `数学`,`动态规划`,`滑动窗口`,`概率与统计` | 中等 | 第 85 场周赛 |
|
||||
| 0838 | [推多米诺](/solution/0800-0899/0838.Push%20Dominoes/README.md) | `双指针`,`字符串`,`动态规划` | 中等 | 第 85 场周赛 |
|
||||
| 0839 | [相似字符串组](/solution/0800-0899/0839.Similar%20String%20Groups/README.md) | `深度优先搜索`,`广度优先搜索`,`并查集`,`数组`,`哈希表`,`字符串` | 困难 | 第 85 场周赛 |
|
||||
| 0840 | [矩阵中的幻方](/solution/0800-0899/0840.Magic%20Squares%20In%20Grid/README.md) | `数组`,`数学`,`矩阵` | 中等 | 第 86 场周赛 |
|
||||
| 0840 | [矩阵中的幻方](/solution/0800-0899/0840.Magic%20Squares%20In%20Grid/README.md) | `数组`,`哈希表`,`数学`,`矩阵` | 中等 | 第 86 场周赛 |
|
||||
| 0841 | [钥匙和房间](/solution/0800-0899/0841.Keys%20and%20Rooms/README.md) | `深度优先搜索`,`广度优先搜索`,`图` | 中等 | 第 86 场周赛 |
|
||||
| 0842 | [将数组拆分成斐波那契序列](/solution/0800-0899/0842.Split%20Array%20into%20Fibonacci%20Sequence/README.md) | `字符串`,`回溯` | 中等 | 第 86 场周赛 |
|
||||
| 0843 | [猜猜这个单词](/solution/0800-0899/0843.Guess%20the%20Word/README.md) | `数组`,`数学`,`字符串`,`博弈`,`交互` | 困难 | 第 86 场周赛 |
|
||||
|
|
@ -2627,7 +2627,7 @@
|
|||
| 2614 | [对角线上的质数](/solution/2600-2699/2614.Prime%20In%20Diagonal/README.md) | `数组`,`数学`,`矩阵`,`数论` | 简单 | 第 340 场周赛 |
|
||||
| 2615 | [等值距离和](/solution/2600-2699/2615.Sum%20of%20Distances/README.md) | `数组`,`哈希表`,`前缀和` | 中等 | 第 340 场周赛 |
|
||||
| 2616 | [最小化数对的最大差值](/solution/2600-2699/2616.Minimize%20the%20Maximum%20Difference%20of%20Pairs/README.md) | `贪心`,`数组`,`二分查找` | 中等 | 第 340 场周赛 |
|
||||
| 2617 | [网格图中最少访问的格子数](/solution/2600-2699/2617.Minimum%20Number%20of%20Visited%20Cells%20in%20a%20Grid/README.md) | `栈`,`并查集`,`树状数组`,`线段树`,`数组`,`二分查找`,`动态规划` | 困难 | 第 340 场周赛 |
|
||||
| 2617 | [网格图中最少访问的格子数](/solution/2600-2699/2617.Minimum%20Number%20of%20Visited%20Cells%20in%20a%20Grid/README.md) | `栈`,`广度优先搜索`,`并查集`,`数组`,`动态规划`,`矩阵`,`单调栈`,`堆（优先队列）` | 困难 | 第 340 场周赛 |
|
||||
| 2618 | [检查是否是类的对象实例](/solution/2600-2699/2618.Check%20if%20Object%20Instance%20of%20Class/README.md) | | 中等 | |
|
||||
| 2619 | [数组原型对象的最后一个元素](/solution/2600-2699/2619.Array%20Prototype%20Last/README.md) | | 简单 | |
|
||||
| 2620 | [计数器](/solution/2600-2699/2620.Counter/README.md) | | 简单 | |
|
||||
|
|
@ -3060,16 +3060,16 @@
|
|||
| 3047 | [求交集区域内的最大正方形面积](/solution/3000-3099/3047.Find%20the%20Largest%20Area%20of%20Square%20Inside%20Two%20Rectangles/README.md) | `几何`,`数组`,`数学` | 中等 | 第 386 场周赛 |
|
||||
| 3048 | [标记所有下标的最早秒数 I](/solution/3000-3099/3048.Earliest%20Second%20to%20Mark%20Indices%20I/README.md) | `数组`,`二分查找` | 中等 | 第 386 场周赛 |
|
||||
| 3049 | [标记所有下标的最早秒数 II](/solution/3000-3099/3049.Earliest%20Second%20to%20Mark%20Indices%20II/README.md) | `贪心`,`数组`,`二分查找`,`堆（优先队列）` | 困难 | 第 386 场周赛 |
|
||||
| 3050 | [Pizza Toppings Cost Analysis](/solution/3000-3099/3050.Pizza%20Toppings%20Cost%20Analysis/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3051 | [Find Candidates for Data Scientist Position](/solution/3000-3099/3051.Find%20Candidates%20for%20Data%20Scientist%20Position/README.md) | `数据库` | 简单 | 🔒 |
|
||||
| 3052 | [Maximize Items](/solution/3000-3099/3052.Maximize%20Items/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 3053 | [Classifying Triangles by Lengths](/solution/3000-3099/3053.Classifying%20Triangles%20by%20Lengths/README.md) | `数据库` | 简单 | 🔒 |
|
||||
| 3054 | [Binary Tree Nodes](/solution/3000-3099/3054.Binary%20Tree%20Nodes/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3055 | [Top Percentile Fraud](/solution/3000-3099/3055.Top%20Percentile%20Fraud/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3056 | [Snaps Analysis](/solution/3000-3099/3056.Snaps%20Analysis/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3057 | [Employees Project Allocation](/solution/3000-3099/3057.Employees%20Project%20Allocation/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 3058 | [Friends With No Mutual Friends](/solution/3000-3099/3058.Friends%20With%20No%20Mutual%20Friends/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3059 | [Find All Unique Email Domains](/solution/3000-3099/3059.Find%20All%20Unique%20Email%20Domains/README.md) | `数据库` | 简单 | 🔒 |
|
||||
| 3050 | [披萨配料成本分析](/solution/3000-3099/3050.Pizza%20Toppings%20Cost%20Analysis/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3051 | [寻找数据科学家职位的候选人](/solution/3000-3099/3051.Find%20Candidates%20for%20Data%20Scientist%20Position/README.md) | `数据库` | 简单 | 🔒 |
|
||||
| 3052 | [最大化商品](/solution/3000-3099/3052.Maximize%20Items/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 3053 | [根据长度分类三角形](/solution/3000-3099/3053.Classifying%20Triangles%20by%20Lengths/README.md) | `数据库` | 简单 | 🔒 |
|
||||
| 3054 | [二叉树节点](/solution/3000-3099/3054.Binary%20Tree%20Nodes/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3055 | [最高欺诈百分位数](/solution/3000-3099/3055.Top%20Percentile%20Fraud/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3056 | [快照分析](/solution/3000-3099/3056.Snaps%20Analysis/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3057 | [员工项目分配](/solution/3000-3099/3057.Employees%20Project%20Allocation/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 3058 | [没有共同朋友的朋友](/solution/3000-3099/3058.Friends%20With%20No%20Mutual%20Friends/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3059 | [找到所有不同的邮件域名](/solution/3000-3099/3059.Find%20All%20Unique%20Email%20Domains/README.md) | `数据库` | 简单 | 🔒 |
|
||||
| 3060 | [时间范围内的用户活动](/solution/3000-3099/3060.User%20Activities%20within%20Time%20Bounds/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 3061 | [计算滞留雨水](/solution/3000-3099/3061.Calculate%20Trapping%20Rain%20Water/README.md) | `数据库` | 困难 | 🔒 |
|
||||
| 3062 | [链表游戏的获胜者](/solution/3000-3099/3062.Winner%20of%20the%20Linked%20List%20Game/README.md) | `链表` | 简单 | 🔒 |
|
||||
|
|
@ -3097,13 +3097,13 @@
|
|||
| 3084 | [统计以给定字符开头和结尾的子字符串总数](/solution/3000-3099/3084.Count%20Substrings%20Starting%20and%20Ending%20with%20Given%20Character/README.md) | `数学`,`字符串`,`计数` | 中等 | 第 389 场周赛 |
|
||||
| 3085 | [成为 K 特殊字符串需要删除的最少字符数](/solution/3000-3099/3085.Minimum%20Deletions%20to%20Make%20String%20K-Special/README.md) | `贪心`,`哈希表`,`字符串`,`计数`,`排序` | 中等 | 第 389 场周赛 |
|
||||
| 3086 | [拾起 K 个 1 需要的最少行动次数](/solution/3000-3099/3086.Minimum%20Moves%20to%20Pick%20K%20Ones/README.md) | `贪心`,`数组`,`前缀和`,`滑动窗口` | 困难 | 第 389 场周赛 |
|
||||
| 3087 | [查找热门话题标签](/solution/3000-3099/3087.Find%20Trending%20Hashtags/README.md) | | 中等 | 🔒 |
|
||||
| 3088 | [使字符串反回文](/solution/3000-3099/3088.Make%20String%20Anti-palindrome/README.md) | | 困难 | 🔒 |
|
||||
| 3089 | [Find Bursty Behavior](/solution/3000-3099/3089.Find%20Bursty%20Behavior/README.md) | | 中等 | 🔒 |
|
||||
| 3090 | [每个字符最多出现两次的最长子字符串](/solution/3000-3099/3090.Maximum%20Length%20Substring%20With%20Two%20Occurrences/README.md) | | 简单 | 第 390 场周赛 |
|
||||
| 3091 | [执行操作使数据元素之和大于等于 K](/solution/3000-3099/3091.Apply%20Operations%20to%20Make%20Sum%20of%20Array%20Greater%20Than%20or%20Equal%20to%20k/README.md) | | 中等 | 第 390 场周赛 |
|
||||
| 3092 | [最高频率的 ID](/solution/3000-3099/3092.Most%20Frequent%20IDs/README.md) | | 中等 | 第 390 场周赛 |
|
||||
| 3093 | [最长公共后缀查询](/solution/3000-3099/3093.Longest%20Common%20Suffix%20Queries/README.md) | | 困难 | 第 390 场周赛 |
|
||||
| 3087 | [查找热门话题标签](/solution/3000-3099/3087.Find%20Trending%20Hashtags/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3088 | [使字符串反回文](/solution/3000-3099/3088.Make%20String%20Anti-palindrome/README.md) | `贪心`,`字符串`,`排序` | 困难 | 🔒 |
|
||||
| 3089 | [查找突发行为](/solution/3000-3099/3089.Find%20Bursty%20Behavior/README.md) | `数据库` | 中等 | 🔒 |
|
||||
| 3090 | [每个字符最多出现两次的最长子字符串](/solution/3000-3099/3090.Maximum%20Length%20Substring%20With%20Two%20Occurrences/README.md) | `哈希表`,`字符串`,`滑动窗口` | 简单 | 第 390 场周赛 |
|
||||
| 3091 | [执行操作使数据元素之和大于等于 K](/solution/3000-3099/3091.Apply%20Operations%20to%20Make%20Sum%20of%20Array%20Greater%20Than%20or%20Equal%20to%20k/README.md) | `贪心`,`数学`,`枚举` | 中等 | 第 390 场周赛 |
|
||||
| 3092 | [最高频率的 ID](/solution/3000-3099/3092.Most%20Frequent%20IDs/README.md) | `数组`,`哈希表`,`有序集合`,`堆（优先队列）` | 中等 | 第 390 场周赛 |
|
||||
| 3093 | [最长公共后缀查询](/solution/3000-3099/3093.Longest%20Common%20Suffix%20Queries/README.md) | `字典树`,`数组`,`字符串` | 困难 | 第 390 场周赛 |
|
||||
|
||||
## 版权
|
||||
|
||||
|
|
|
|||
|
|
@ -329,7 +329,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
|
|||
| 0318 | [Maximum Product of Word Lengths](/solution/0300-0399/0318.Maximum%20Product%20of%20Word%20Lengths/README_EN.md) | `Bit Manipulation`,`Array`,`String` | Medium | |
|
||||
| 0319 | [Bulb Switcher](/solution/0300-0399/0319.Bulb%20Switcher/README_EN.md) | `Brainteaser`,`Math` | Medium | |
|
||||
| 0320 | [Generalized Abbreviation](/solution/0300-0399/0320.Generalized%20Abbreviation/README_EN.md) | `Bit Manipulation`,`String`,`Backtracking` | Medium | 🔒 |
|
||||
| 0321 | [Create Maximum Number](/solution/0300-0399/0321.Create%20Maximum%20Number/README_EN.md) | `Stack`,`Greedy`,`Monotonic Stack` | Hard | |
|
||||
| 0321 | [Create Maximum Number](/solution/0300-0399/0321.Create%20Maximum%20Number/README_EN.md) | `Stack`,`Greedy`,`Array`,`Two Pointers`,`Monotonic Stack` | Hard | |
|
||||
| 0322 | [Coin Change](/solution/0300-0399/0322.Coin%20Change/README_EN.md) | `Breadth-First Search`,`Array`,`Dynamic Programming` | Medium | |
|
||||
| 0323 | [Number of Connected Components in an Undirected Graph](/solution/0300-0399/0323.Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph/README_EN.md) | `Depth-First Search`,`Breadth-First Search`,`Union Find`,`Graph` | Medium | 🔒 |
|
||||
| 0324 | [Wiggle Sort II](/solution/0300-0399/0324.Wiggle%20Sort%20II/README_EN.md) | `Array`,`Divide and Conquer`,`Quickselect`,`Sorting` | Medium | |
|
||||
|
|
@ -848,7 +848,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
|
|||
| 0837 | [New 21 Game](/solution/0800-0899/0837.New%2021%20Game/README_EN.md) | `Math`,`Dynamic Programming`,`Sliding Window`,`Probability and Statistics` | Medium | Weekly Contest 85 |
|
||||
| 0838 | [Push Dominoes](/solution/0800-0899/0838.Push%20Dominoes/README_EN.md) | `Two Pointers`,`String`,`Dynamic Programming` | Medium | Weekly Contest 85 |
|
||||
| 0839 | [Similar String Groups](/solution/0800-0899/0839.Similar%20String%20Groups/README_EN.md) | `Depth-First Search`,`Breadth-First Search`,`Union Find`,`Array`,`Hash Table`,`String` | Hard | Weekly Contest 85 |
|
||||
| 0840 | [Magic Squares In Grid](/solution/0800-0899/0840.Magic%20Squares%20In%20Grid/README_EN.md) | `Array`,`Math`,`Matrix` | Medium | Weekly Contest 86 |
|
||||
| 0840 | [Magic Squares In Grid](/solution/0800-0899/0840.Magic%20Squares%20In%20Grid/README_EN.md) | `Array`,`Hash Table`,`Math`,`Matrix` | Medium | Weekly Contest 86 |
|
||||
| 0841 | [Keys and Rooms](/solution/0800-0899/0841.Keys%20and%20Rooms/README_EN.md) | `Depth-First Search`,`Breadth-First Search`,`Graph` | Medium | Weekly Contest 86 |
|
||||
| 0842 | [Split Array into Fibonacci Sequence](/solution/0800-0899/0842.Split%20Array%20into%20Fibonacci%20Sequence/README_EN.md) | `String`,`Backtracking` | Medium | Weekly Contest 86 |
|
||||
| 0843 | [Guess the Word](/solution/0800-0899/0843.Guess%20the%20Word/README_EN.md) | `Array`,`Math`,`String`,`Game Theory`,`Interactive` | Hard | Weekly Contest 86 |
|
||||
|
|
@ -2625,7 +2625,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
|
|||
| 2614 | [Prime In Diagonal](/solution/2600-2699/2614.Prime%20In%20Diagonal/README_EN.md) | `Array`,`Math`,`Matrix`,`Number Theory` | Easy | Weekly Contest 340 |
|
||||
| 2615 | [Sum of Distances](/solution/2600-2699/2615.Sum%20of%20Distances/README_EN.md) | `Array`,`Hash Table`,`Prefix Sum` | Medium | Weekly Contest 340 |
|
||||
| 2616 | [Minimize the Maximum Difference of Pairs](/solution/2600-2699/2616.Minimize%20the%20Maximum%20Difference%20of%20Pairs/README_EN.md) | `Greedy`,`Array`,`Binary Search` | Medium | Weekly Contest 340 |
|
||||
| 2617 | [Minimum Number of Visited Cells in a Grid](/solution/2600-2699/2617.Minimum%20Number%20of%20Visited%20Cells%20in%20a%20Grid/README_EN.md) | `Stack`,`Union Find`,`Binary Indexed Tree`,`Segment Tree`,`Array`,`Binary Search`,`Dynamic Programming` | Hard | Weekly Contest 340 |
|
||||
| 2617 | [Minimum Number of Visited Cells in a Grid](/solution/2600-2699/2617.Minimum%20Number%20of%20Visited%20Cells%20in%20a%20Grid/README_EN.md) | `Stack`,`Breadth-First Search`,`Union Find`,`Array`,`Dynamic Programming`,`Matrix`,`Monotonic Stack`,`Heap (Priority Queue)` | Hard | Weekly Contest 340 |
|
||||
| 2618 | [Check if Object Instance of Class](/solution/2600-2699/2618.Check%20if%20Object%20Instance%20of%20Class/README_EN.md) | | Medium | |
|
||||
| 2619 | [Array Prototype Last](/solution/2600-2699/2619.Array%20Prototype%20Last/README_EN.md) | | Easy | |
|
||||
| 2620 | [Counter](/solution/2600-2699/2620.Counter/README_EN.md) | | Easy | |
|
||||
|
|
@ -3095,13 +3095,13 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
|
|||
| 3084 | [Count Substrings Starting and Ending with Given Character](/solution/3000-3099/3084.Count%20Substrings%20Starting%20and%20Ending%20with%20Given%20Character/README_EN.md) | `Math`,`String`,`Counting` | Medium | Weekly Contest 389 |
|
||||
| 3085 | [Minimum Deletions to Make String K-Special](/solution/3000-3099/3085.Minimum%20Deletions%20to%20Make%20String%20K-Special/README_EN.md) | `Greedy`,`Hash Table`,`String`,`Counting`,`Sorting` | Medium | Weekly Contest 389 |
|
||||
| 3086 | [Minimum Moves to Pick K Ones](/solution/3000-3099/3086.Minimum%20Moves%20to%20Pick%20K%20Ones/README_EN.md) | `Greedy`,`Array`,`Prefix Sum`,`Sliding Window` | Hard | Weekly Contest 389 |
|
||||
| 3087 | [Find Trending Hashtags](/solution/3000-3099/3087.Find%20Trending%20Hashtags/README_EN.md) | | Medium | 🔒 |
|
||||
| 3088 | [Make String Anti-palindrome](/solution/3000-3099/3088.Make%20String%20Anti-palindrome/README_EN.md) | | Hard | 🔒 |
|
||||
| 3089 | [Find Bursty Behavior](/solution/3000-3099/3089.Find%20Bursty%20Behavior/README_EN.md) | | Medium | 🔒 |
|
||||
| 3090 | [Maximum Length Substring With Two Occurrences](/solution/3000-3099/3090.Maximum%20Length%20Substring%20With%20Two%20Occurrences/README_EN.md) | | Easy | Weekly Contest 390 |
|
||||
| 3091 | [Apply Operations to Make Sum of Array Greater Than or Equal to k](/solution/3000-3099/3091.Apply%20Operations%20to%20Make%20Sum%20of%20Array%20Greater%20Than%20or%20Equal%20to%20k/README_EN.md) | | Medium | Weekly Contest 390 |
|
||||
| 3092 | [Most Frequent IDs](/solution/3000-3099/3092.Most%20Frequent%20IDs/README_EN.md) | | Medium | Weekly Contest 390 |
|
||||
| 3093 | [Longest Common Suffix Queries](/solution/3000-3099/3093.Longest%20Common%20Suffix%20Queries/README_EN.md) | | Hard | Weekly Contest 390 |
|
||||
| 3087 | [Find Trending Hashtags](/solution/3000-3099/3087.Find%20Trending%20Hashtags/README_EN.md) | `Database` | Medium | 🔒 |
|
||||
| 3088 | [Make String Anti-palindrome](/solution/3000-3099/3088.Make%20String%20Anti-palindrome/README_EN.md) | `Greedy`,`String`,`Sorting` | Hard | 🔒 |
|
||||
| 3089 | [Find Bursty Behavior](/solution/3000-3099/3089.Find%20Bursty%20Behavior/README_EN.md) | `Database` | Medium | 🔒 |
|
||||
| 3090 | [Maximum Length Substring With Two Occurrences](/solution/3000-3099/3090.Maximum%20Length%20Substring%20With%20Two%20Occurrences/README_EN.md) | `Hash Table`,`String`,`Sliding Window` | Easy | Weekly Contest 390 |
|
||||
| 3091 | [Apply Operations to Make Sum of Array Greater Than or Equal to k](/solution/3000-3099/3091.Apply%20Operations%20to%20Make%20Sum%20of%20Array%20Greater%20Than%20or%20Equal%20to%20k/README_EN.md) | `Greedy`,`Math`,`Enumeration` | Medium | Weekly Contest 390 |
|
||||
| 3092 | [Most Frequent IDs](/solution/3000-3099/3092.Most%20Frequent%20IDs/README_EN.md) | `Array`,`Hash Table`,`Ordered Set`,`Heap (Priority Queue)` | Medium | Weekly Contest 390 |
|
||||
| 3093 | [Longest Common Suffix Queries](/solution/3000-3099/3093.Longest%20Common%20Suffix%20Queries/README_EN.md) | `Trie`,`Array`,`String` | Hard | Weekly Contest 390 |
|
||||
|
||||
## Copyright
|
||||
|
||||
|
|
|
|||
File diff suppressed because one or more lines are too long
|
|
@ -249,17 +249,17 @@
|
|||
- [2993.发生在周五的交易 I](/database-solution/2900-2999/2993.Friday%20Purchases%20I/README.md)
|
||||
- [2994.发生在周五的交易 II](/database-solution/2900-2999/2994.Friday%20Purchases%20II/README.md)
|
||||
- [2995.观众变主播](/database-solution/2900-2999/2995.Viewers%20Turned%20Streamers/README.md)
|
||||
- [3050.Pizza Toppings Cost Analysis](/database-solution/3000-3099/3050.Pizza%20Toppings%20Cost%20Analysis/README.md)
|
||||
- [3051.Find Candidates for Data Scientist Position](/database-solution/3000-3099/3051.Find%20Candidates%20for%20Data%20Scientist%20Position/README.md)
|
||||
- [3052.Maximize Items](/database-solution/3000-3099/3052.Maximize%20Items/README.md)
|
||||
- [3053.Classifying Triangles by Lengths](/database-solution/3000-3099/3053.Classifying%20Triangles%20by%20Lengths/README.md)
|
||||
- [3054.Binary Tree Nodes](/database-solution/3000-3099/3054.Binary%20Tree%20Nodes/README.md)
|
||||
- [3055.Top Percentile Fraud](/database-solution/3000-3099/3055.Top%20Percentile%20Fraud/README.md)
|
||||
- [3056.Snaps Analysis](/database-solution/3000-3099/3056.Snaps%20Analysis/README.md)
|
||||
- [3057.Employees Project Allocation](/database-solution/3000-3099/3057.Employees%20Project%20Allocation/README.md)
|
||||
- [3058.Friends With No Mutual Friends](/database-solution/3000-3099/3058.Friends%20With%20No%20Mutual%20Friends/README.md)
|
||||
- [3059.Find All Unique Email Domains](/database-solution/3000-3099/3059.Find%20All%20Unique%20Email%20Domains/README.md)
|
||||
- [3050.披萨配料成本分析](/database-solution/3000-3099/3050.Pizza%20Toppings%20Cost%20Analysis/README.md)
|
||||
- [3051.寻找数据科学家职位的候选人](/database-solution/3000-3099/3051.Find%20Candidates%20for%20Data%20Scientist%20Position/README.md)
|
||||
- [3052.最大化商品](/database-solution/3000-3099/3052.Maximize%20Items/README.md)
|
||||
- [3053.根据长度分类三角形](/database-solution/3000-3099/3053.Classifying%20Triangles%20by%20Lengths/README.md)
|
||||
- [3054.二叉树节点](/database-solution/3000-3099/3054.Binary%20Tree%20Nodes/README.md)
|
||||
- [3055.最高欺诈百分位数](/database-solution/3000-3099/3055.Top%20Percentile%20Fraud/README.md)
|
||||
- [3056.快照分析](/database-solution/3000-3099/3056.Snaps%20Analysis/README.md)
|
||||
- [3057.员工项目分配](/database-solution/3000-3099/3057.Employees%20Project%20Allocation/README.md)
|
||||
- [3058.没有共同朋友的朋友](/database-solution/3000-3099/3058.Friends%20With%20No%20Mutual%20Friends/README.md)
|
||||
- [3059.找到所有不同的邮件域名](/database-solution/3000-3099/3059.Find%20All%20Unique%20Email%20Domains/README.md)
|
||||
- [3060.时间范围内的用户活动](/database-solution/3000-3099/3060.User%20Activities%20within%20Time%20Bounds/README.md)
|
||||
- [3061.计算滞留雨水](/database-solution/3000-3099/3061.Calculate%20Trapping%20Rain%20Water/README.md)
|
||||
- [3087.查找热门话题标签](/database-solution/3000-3099/3087.Find%20Trending%20Hashtags/README.md)
|
||||
- [3089.Find Bursty Behavior](/database-solution/3000-3099/3089.Find%20Bursty%20Behavior/README.md)
|
||||
- [3089.查找突发行为](/database-solution/3000-3099/3089.Find%20Bursty%20Behavior/README.md)
|
||||
|
|
|
|||
|
|
@ -144,7 +144,7 @@ This work is licensed under a <a rel="license" href="http://creativecommons.org/
|
|||
|
||||
## 赛后估分网站
|
||||
|
||||
- https://lccn.lbao.site
|
||||
如果你想在比赛结束后估算自己的积分变化，可以访问网站 [LeetCode Contest Rating Predictor](https://lccn.lbao.site/)。
|
||||
|
||||
## 往期竞赛
|
||||
|
||||
|
|
@ -179,7 +179,7 @@ For top 10 users (excluding LCCN users), your LeetCode ID will be colored orange
|
|||
|
||||
## Rating Predictor
|
||||
|
||||
Get your rating changes right after the completion of LeetCode contests, https://lccn.lbao.site
|
||||
If you want to estimate your score changes after the contest ends, you can visit the website [LeetCode Contest Rating Predictor](https://lccn.lbao.site/).
|
||||
|
||||
## Past Contests
|
||||
|
||||
|
|
|
|||
Loading…
Reference in New Issue