mirror of https://github.com/doocs/leetcode.git
feat: add solutions to lc problem: No.0734 (#4025)
No.0734.Sentence Similarity
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Libin YANG
GitHub
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commit
b901bf832b
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@ -63,7 +63,13 @@ myCalendar.book(20, 30); // return True, The event can be booked, as the first e
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Ordered Set
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We can use an ordered set to store the schedule. An ordered set can perform insert, delete, and search operations in $O(\log n)$ time. The elements in the ordered set are sorted by the $\textit{endTime}$ of the schedule in ascending order.
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When calling the $\text{book}(start, end)$ method, we search for the first schedule in the ordered set with an end time greater than $\textit{start}$. If it exists and its start time is less than $\textit{end}$, it means there is a double booking, and we return $\text{false}$. Otherwise, we insert $\textit{end}$ as the key and $\textit{start}$ as the value into the ordered set and return $\text{true}$.
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The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the number of schedules.
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<!-- tabs:start -->
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@ -80,6 +80,16 @@ tags:
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### 方法一：哈希表
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我们首先判断 $\textit{sentence1}$ 和 $\textit{sentence2}$ 的长度是否相等，如果不相等则返回 $\text{false}$。
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然后我们使用一个哈希表 $\textit{s}$ 来存储所有相似的单词对，对于 $\textit{similarPairs}$ 中的每一个单词对 $[x, y]$，我们将 $x$ 和 $y$ 加入到哈希表 $\textit{s}$ 中。
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接下来我们遍历 $\textit{sentence1}$ 和 $\textit{sentence2}$，对于每一个位置 $i$，如果 $\textit{sentence1}[i]$ 不等于 $\textit{sentence2}[i]$，并且 $(\textit{sentence1}[i], \textit{sentence2}[i])$ 和 $(\textit{sentence2}[i], \textit{sentence1}[i])$ 都不在哈希表 $\textit{s}$ 中，那么返回 $\text{false}$。
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如果遍历结束后都没有返回 $\text{false}$，说明 $\textit{sentence1}$ 和 $\textit{sentence2}$ 是相似的，返回 $\text{true}$。
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时间复杂度 $O(L)$，空间复杂度 $O(L)$，其中 $L$ 为题目中所有字符串的长度之和。
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<!-- tabs:start -->
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#### Python3
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@ -91,10 +101,11 @@ class Solution:
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) -> bool:
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if len(sentence1) != len(sentence2):
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return False
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s = {(a, b) for a, b in similarPairs}
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return all(
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a == b or (a, b) in s or (b, a) in s for a, b in zip(sentence1, sentence2)
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)
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s = {(x, y) for x, y in similarPairs}
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for x, y in zip(sentence1, sentence2):
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if x != y and (x, y) not in s and (y, x) not in s:
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return False
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return True
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```
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#### Java
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@ -106,13 +117,14 @@ class Solution {
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if (sentence1.length != sentence2.length) {
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return false;
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}
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Set<String> s = new HashSet<>();
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for (List<String> e : similarPairs) {
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s.add(e.get(0) + "." + e.get(1));
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Set<List<String>> s = new HashSet<>();
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for (var p : similarPairs) {
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s.add(p);
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}
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for (int i = 0; i < sentence1.length; ++i) {
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String a = sentence1[i], b = sentence2[i];
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if (!a.equals(b) && !s.contains(a + "." + b) && !s.contains(b + "." + a)) {
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for (int i = 0; i < sentence1.length; i++) {
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if (!sentence1[i].equals(sentence2[i])
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&& !s.contains(List.of(sentence1[i], sentence2[i]))
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&& !s.contains(List.of(sentence2[i], sentence1[i]))) {
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return false;
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}
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}
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@ -127,13 +139,18 @@ class Solution {
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class Solution {
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public:
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bool areSentencesSimilar(vector<string>& sentence1, vector<string>& sentence2, vector<vector<string>>& similarPairs) {
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int m = sentence1.size(), n = sentence2.size();
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if (m != n) return false;
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if (sentence1.size() != sentence2.size()) {
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return false;
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}
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unordered_set<string> s;
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for (auto e : similarPairs) s.insert(e[0] + "." + e[1]);
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for (int i = 0; i < n; ++i) {
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string a = sentence1[i], b = sentence2[i];
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if (a != b && !s.count(a + "." + b) && !s.count(b + "." + a)) return false;
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for (const auto& p : similarPairs) {
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s.insert(p[0] + "#" + p[1]);
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s.insert(p[1] + "#" + p[0]);
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}
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for (int i = 0; i < sentence1.size(); ++i) {
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if (sentence1[i] != sentence2[i] && !s.contains(sentence1[i] + "#" + sentence2[i])) {
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return false;
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}
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}
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return true;
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}
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@ -148,12 +165,12 @@ func areSentencesSimilar(sentence1 []string, sentence2 []string, similarPairs []
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return false
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}
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s := map[string]bool{}
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for _, e := range similarPairs {
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s[e[0]+"."+e[1]] = true
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for _, p := range similarPairs {
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s[p[0]+"#"+p[1]] = true
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}
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for i, a := range sentence1 {
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b := sentence2[i]
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if a != b && !s[a+"."+b] && !s[b+"."+a] {
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for i, x := range sentence1 {
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y := sentence2[i]
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if x != y && !s[x+"#"+y] && !s[y+"#"+x] {
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return false
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}
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}
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@ -161,6 +178,91 @@ func areSentencesSimilar(sentence1 []string, sentence2 []string, similarPairs []
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}
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```
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#### TypeScript
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```ts
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function areSentencesSimilar(
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sentence1: string[],
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sentence2: string[],
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similarPairs: string[][],
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): boolean {
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if (sentence1.length !== sentence2.length) {
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return false;
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}
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const s = new Set<string>();
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for (const [x, y] of similarPairs) {
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s.add(x + '#' + y);
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s.add(y + '#' + x);
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}
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for (let i = 0; i < sentence1.length; i++) {
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if (sentence1[i] !== sentence2[i] && !s.has(sentence1[i] + '#' + sentence2[i])) {
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return false;
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}
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}
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return true;
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}
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```
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#### Rust
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```rust
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use std::collections::HashSet;
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impl Solution {
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pub fn are_sentences_similar(
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sentence1: Vec<String>,
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sentence2: Vec<String>,
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similar_pairs: Vec<Vec<String>>,
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) -> bool {
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if sentence1.len() != sentence2.len() {
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return false;
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}
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let s: HashSet<(String, String)> = similar_pairs
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.into_iter()
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.map(|pair| (pair[0].clone(), pair[1].clone()))
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.collect();
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for (x, y) in sentence1.iter().zip(sentence2.iter()) {
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if x != y
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&& !s.contains(&(x.clone(), y.clone()))
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&& !s.contains(&(y.clone(), x.clone()))
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{
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return false;
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}
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}
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true
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}
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}
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```
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#### JavaScript
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```js
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/**
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* @param {string[]} sentence1
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* @param {string[]} sentence2
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* @param {string[][]} similarPairs
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* @return {boolean}
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*/
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var areSentencesSimilar = function (sentence1, sentence2, similarPairs) {
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if (sentence1.length !== sentence2.length) {
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return false;
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}
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const s = new Set();
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for (const [x, y] of similarPairs) {
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s.add(x + '#' + y);
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s.add(y + '#' + x);
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}
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for (let i = 0; i < sentence1.length; i++) {
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if (sentence1[i] !== sentence2[i] && !s.has(sentence1[i] + '#' + sentence2[i])) {
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return false;
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}
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}
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return true;
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};
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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@ -78,7 +78,17 @@ tags:
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Hash Table
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First, we check if the lengths of $\textit{sentence1}$ and $\textit{sentence2}$ are equal. If they are not equal, return $\text{false}$.
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Then we use a hash table $\textit{s}$ to store all similar word pairs. For each word pair $[x, y]$ in $\textit{similarPairs}$, we add $x$ and $y$ to the hash table $\textit{s}$.
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Next, we traverse $\textit{sentence1}$ and $\textit{sentence2}$. For each position $i$, if $\textit{sentence1}[i]$ is not equal to $\textit{sentence2}[i]$, and $(\textit{sentence1}[i], \textit{sentence2}[i])$ and $(\textit{sentence2}[i], \textit{sentence1}[i])$ are not in the hash table $\textit{s}$, then return $\text{false}$.
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If the traversal ends without returning $\text{false}$, it means $\textit{sentence1}$ and $\textit{sentence2}$ are similar, so return $\text{true}$.
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The time complexity is $O(L)$, and the space complexity is $O(L)$, where $L$ is the sum of the lengths of all strings in the problem.
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<!-- tabs:start -->
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@ -91,10 +101,11 @@ class Solution:
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) -> bool:
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if len(sentence1) != len(sentence2):
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return False
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s = {(a, b) for a, b in similarPairs}
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return all(
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a == b or (a, b) in s or (b, a) in s for a, b in zip(sentence1, sentence2)
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)
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s = {(x, y) for x, y in similarPairs}
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for x, y in zip(sentence1, sentence2):
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if x != y and (x, y) not in s and (y, x) not in s:
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return False
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return True
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```
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#### Java
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@ -106,13 +117,14 @@ class Solution {
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if (sentence1.length != sentence2.length) {
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return false;
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}
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Set<String> s = new HashSet<>();
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for (List<String> e : similarPairs) {
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s.add(e.get(0) + "." + e.get(1));
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Set<List<String>> s = new HashSet<>();
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for (var p : similarPairs) {
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s.add(p);
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}
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for (int i = 0; i < sentence1.length; ++i) {
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String a = sentence1[i], b = sentence2[i];
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if (!a.equals(b) && !s.contains(a + "." + b) && !s.contains(b + "." + a)) {
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for (int i = 0; i < sentence1.length; i++) {
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if (!sentence1[i].equals(sentence2[i])
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&& !s.contains(List.of(sentence1[i], sentence2[i]))
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&& !s.contains(List.of(sentence2[i], sentence1[i]))) {
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return false;
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}
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}
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@ -127,13 +139,18 @@ class Solution {
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class Solution {
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public:
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bool areSentencesSimilar(vector<string>& sentence1, vector<string>& sentence2, vector<vector<string>>& similarPairs) {
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int m = sentence1.size(), n = sentence2.size();
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if (m != n) return false;
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if (sentence1.size() != sentence2.size()) {
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return false;
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}
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unordered_set<string> s;
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for (auto e : similarPairs) s.insert(e[0] + "." + e[1]);
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for (int i = 0; i < n; ++i) {
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string a = sentence1[i], b = sentence2[i];
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if (a != b && !s.count(a + "." + b) && !s.count(b + "." + a)) return false;
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for (const auto& p : similarPairs) {
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s.insert(p[0] + "#" + p[1]);
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s.insert(p[1] + "#" + p[0]);
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}
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for (int i = 0; i < sentence1.size(); ++i) {
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if (sentence1[i] != sentence2[i] && !s.contains(sentence1[i] + "#" + sentence2[i])) {
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return false;
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}
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}
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return true;
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}
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@ -148,12 +165,12 @@ func areSentencesSimilar(sentence1 []string, sentence2 []string, similarPairs []
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return false
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}
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s := map[string]bool{}
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for _, e := range similarPairs {
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s[e[0]+"."+e[1]] = true
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for _, p := range similarPairs {
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s[p[0]+"#"+p[1]] = true
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}
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for i, a := range sentence1 {
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b := sentence2[i]
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if a != b && !s[a+"."+b] && !s[b+"."+a] {
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for i, x := range sentence1 {
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y := sentence2[i]
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if x != y && !s[x+"#"+y] && !s[y+"#"+x] {
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return false
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}
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}
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@ -161,6 +178,91 @@ func areSentencesSimilar(sentence1 []string, sentence2 []string, similarPairs []
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}
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```
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#### TypeScript
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```ts
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function areSentencesSimilar(
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sentence1: string[],
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sentence2: string[],
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similarPairs: string[][],
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): boolean {
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if (sentence1.length !== sentence2.length) {
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return false;
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}
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const s = new Set<string>();
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for (const [x, y] of similarPairs) {
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s.add(x + '#' + y);
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s.add(y + '#' + x);
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}
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for (let i = 0; i < sentence1.length; i++) {
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if (sentence1[i] !== sentence2[i] && !s.has(sentence1[i] + '#' + sentence2[i])) {
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return false;
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}
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}
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return true;
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}
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```
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#### Rust
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```rust
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use std::collections::HashSet;
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impl Solution {
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pub fn are_sentences_similar(
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sentence1: Vec<String>,
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sentence2: Vec<String>,
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similar_pairs: Vec<Vec<String>>,
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) -> bool {
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if sentence1.len() != sentence2.len() {
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return false;
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}
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let s: HashSet<(String, String)> = similar_pairs
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.into_iter()
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.map(|pair| (pair[0].clone(), pair[1].clone()))
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.collect();
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for (x, y) in sentence1.iter().zip(sentence2.iter()) {
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if x != y
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&& !s.contains(&(x.clone(), y.clone()))
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&& !s.contains(&(y.clone(), x.clone()))
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{
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return false;
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}
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}
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true
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}
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}
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```
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#### JavaScript
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```js
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/**
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* @param {string[]} sentence1
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* @param {string[]} sentence2
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* @param {string[][]} similarPairs
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* @return {boolean}
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*/
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var areSentencesSimilar = function (sentence1, sentence2, similarPairs) {
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if (sentence1.length !== sentence2.length) {
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return false;
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}
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const s = new Set();
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for (const [x, y] of similarPairs) {
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s.add(x + '#' + y);
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s.add(y + '#' + x);
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}
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for (let i = 0; i < sentence1.length; i++) {
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if (sentence1[i] !== sentence2[i] && !s.has(sentence1[i] + '#' + sentence2[i])) {
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return false;
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}
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}
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return true;
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};
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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|
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@ -1,14 +1,19 @@
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class Solution {
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public:
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bool areSentencesSimilar(vector<string>& sentence1, vector<string>& sentence2, vector<vector<string>>& similarPairs) {
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int m = sentence1.size(), n = sentence2.size();
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if (m != n) return false;
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if (sentence1.size() != sentence2.size()) {
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return false;
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}
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unordered_set<string> s;
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for (auto e : similarPairs) s.insert(e[0] + "." + e[1]);
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for (int i = 0; i < n; ++i) {
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string a = sentence1[i], b = sentence2[i];
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if (a != b && !s.count(a + "." + b) && !s.count(b + "." + a)) return false;
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for (const auto& p : similarPairs) {
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s.insert(p[0] + "#" + p[1]);
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s.insert(p[1] + "#" + p[0]);
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}
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for (int i = 0; i < sentence1.size(); ++i) {
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if (sentence1[i] != sentence2[i] && !s.contains(sentence1[i] + "#" + sentence2[i])) {
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return false;
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}
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}
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return true;
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}
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};
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};
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|
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@ -3,14 +3,14 @@ func areSentencesSimilar(sentence1 []string, sentence2 []string, similarPairs []
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return false
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}
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s := map[string]bool{}
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for _, e := range similarPairs {
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s[e[0]+"."+e[1]] = true
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for _, p := range similarPairs {
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s[p[0]+"#"+p[1]] = true
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}
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for i, a := range sentence1 {
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b := sentence2[i]
|
||||
if a != b && !s[a+"."+b] && !s[b+"."+a] {
|
||||
for i, x := range sentence1 {
|
||||
y := sentence2[i]
|
||||
if x != y && !s[x+"#"+y] && !s[y+"#"+x] {
|
||||
return false
|
||||
}
|
||||
}
|
||||
return true
|
||||
}
|
||||
}
|
||||
|
|
|
|||
|
|
@ -4,16 +4,17 @@ class Solution {
|
|||
if (sentence1.length != sentence2.length) {
|
||||
return false;
|
||||
}
|
||||
Set<String> s = new HashSet<>();
|
||||
for (List<String> e : similarPairs) {
|
||||
s.add(e.get(0) + "." + e.get(1));
|
||||
Set<List<String>> s = new HashSet<>();
|
||||
for (var p : similarPairs) {
|
||||
s.add(p);
|
||||
}
|
||||
for (int i = 0; i < sentence1.length; ++i) {
|
||||
String a = sentence1[i], b = sentence2[i];
|
||||
if (!a.equals(b) && !s.contains(a + "." + b) && !s.contains(b + "." + a)) {
|
||||
for (int i = 0; i < sentence1.length; i++) {
|
||||
if (!sentence1[i].equals(sentence2[i])
|
||||
&& !s.contains(List.of(sentence1[i], sentence2[i]))
|
||||
&& !s.contains(List.of(sentence2[i], sentence1[i]))) {
|
||||
return false;
|
||||
}
|
||||
}
|
||||
return true;
|
||||
}
|
||||
}
|
||||
}
|
||||
|
|
|
|||
|
|
@ -0,0 +1,22 @@
|
|||
/**
|
||||
* @param {string[]} sentence1
|
||||
* @param {string[]} sentence2
|
||||
* @param {string[][]} similarPairs
|
||||
* @return {boolean}
|
||||
*/
|
||||
var areSentencesSimilar = function (sentence1, sentence2, similarPairs) {
|
||||
if (sentence1.length !== sentence2.length) {
|
||||
return false;
|
||||
}
|
||||
const s = new Set();
|
||||
for (const [x, y] of similarPairs) {
|
||||
s.add(x + '#' + y);
|
||||
s.add(y + '#' + x);
|
||||
}
|
||||
for (let i = 0; i < sentence1.length; i++) {
|
||||
if (sentence1[i] !== sentence2[i] && !s.has(sentence1[i] + '#' + sentence2[i])) {
|
||||
return false;
|
||||
}
|
||||
}
|
||||
return true;
|
||||
};
|
||||
|
|
@ -4,7 +4,8 @@ class Solution:
|
|||
) -> bool:
|
||||
if len(sentence1) != len(sentence2):
|
||||
return False
|
||||
s = {(a, b) for a, b in similarPairs}
|
||||
return all(
|
||||
a == b or (a, b) in s or (b, a) in s for a, b in zip(sentence1, sentence2)
|
||||
)
|
||||
s = {(x, y) for x, y in similarPairs}
|
||||
for x, y in zip(sentence1, sentence2):
|
||||
if x != y and (x, y) not in s and (y, x) not in s:
|
||||
return False
|
||||
return True
|
||||
|
|
|
|||
|
|
@ -0,0 +1,28 @@
|
|||
use std::collections::HashSet;
|
||||
|
||||
impl Solution {
|
||||
pub fn are_sentences_similar(
|
||||
sentence1: Vec<String>,
|
||||
sentence2: Vec<String>,
|
||||
similar_pairs: Vec<Vec<String>>,
|
||||
) -> bool {
|
||||
if sentence1.len() != sentence2.len() {
|
||||
return false;
|
||||
}
|
||||
|
||||
let s: HashSet<(String, String)> = similar_pairs
|
||||
.into_iter()
|
||||
.map(|pair| (pair[0].clone(), pair[1].clone()))
|
||||
.collect();
|
||||
|
||||
for (x, y) in sentence1.iter().zip(sentence2.iter()) {
|
||||
if x != y
|
||||
&& !s.contains(&(x.clone(), y.clone()))
|
||||
&& !s.contains(&(y.clone(), x.clone()))
|
||||
{
|
||||
return false;
|
||||
}
|
||||
}
|
||||
true
|
||||
}
|
||||
}
|
||||
|
|
@ -0,0 +1,20 @@
|
|||
function areSentencesSimilar(
|
||||
sentence1: string[],
|
||||
sentence2: string[],
|
||||
similarPairs: string[][],
|
||||
): boolean {
|
||||
if (sentence1.length !== sentence2.length) {
|
||||
return false;
|
||||
}
|
||||
const s = new Set<string>();
|
||||
for (const [x, y] of similarPairs) {
|
||||
s.add(x + '#' + y);
|
||||
s.add(y + '#' + x);
|
||||
}
|
||||
for (let i = 0; i < sentence1.length; i++) {
|
||||
if (sentence1[i] !== sentence2[i] && !s.has(sentence1[i] + '#' + sentence2[i])) {
|
||||
return false;
|
||||
}
|
||||
}
|
||||
return true;
|
||||
}
|
||||
Loading…
Reference in New Issue