mirror of https://github.com/doocs/leetcode.git
feat: add solutions to lc problem: No.0985 (#4131)
No.0985.Sum of Even Numbers After Queries
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@ -59,15 +59,11 @@ tags:
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### 方法一：模拟
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我们用一个变量 $s$ 记录数组 $nums$ 中所有偶数的和。
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我们用一个整型变量 $\textit{s}$ 记录数组 $\textit{nums}$ 中所有偶数的和，初始时 $\textit{s}$ 为数组 $\textit{nums}$ 中所有偶数的和。
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对于每次查询 $(v, i)$：
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对于每次查询 $(v, i)$，我们首先判断 $\textit{nums}[i]$ 是否为偶数，若 $\textit{nums}[i]$ 为偶数，则将 $\textit{s}$ 减去 $\textit{nums}[i]$；然后将 $\textit{nums}[i]$ 加上 $v$；若 $\textit{nums}[i]$ 为偶数，则将 $\textit{s}$ 加上 $\textit{nums}[i]$，然后将 $\textit{s}$ 加入答案数组。
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我们先判断 $nums[i]$ 是否为偶数，若 $nums[i]$ 为偶数，则将 $s$ 减去 $nums[i]$；然后将 $nums[i]$ 加上 $v$；
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若 $nums[i]$ 为偶数，则将 $s$ 加上 $nums[i]$，然后将 $s$ 加入答案数组。
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时间复杂度 $O(n + m)$，其中 $n$ 和 $m$ 分别为数组 $nums$ 和 $queries$ 的长度。忽略答案数组的空间消耗，空间复杂度 $O(1)$。
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时间复杂度 $O(n + m)$，其中 $n$ 和 $m$ 分别为数组 $\textit{nums}$ 和 $\textit{queries}$ 的长度。忽略答案数组的空间消耗，空间复杂度 $O(1)$。
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<!-- tabs:start -->
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@ -178,12 +174,7 @@ func sumEvenAfterQueries(nums []int, queries [][]int) (ans []int) {
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```ts
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function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
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let s = 0;
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for (const x of nums) {
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if (x % 2 === 0) {
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s += x;
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}
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}
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let s = nums.reduce((acc, x) => acc + (x % 2 === 0 ? x : 0), 0);
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const ans: number[] = [];
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for (const [v, i] of queries) {
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if (nums[i] % 2 === 0) {
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@ -199,6 +190,31 @@ function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn sum_even_after_queries(mut nums: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
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let mut s: i32 = nums.iter().filter(|&x| x % 2 == 0).sum();
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let mut ans = Vec::with_capacity(queries.len());
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for query in queries {
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let (v, i) = (query[0], query[1] as usize);
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if nums[i] % 2 == 0 {
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s -= nums[i];
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}
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nums[i] += v;
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if nums[i] % 2 == 0 {
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s += nums[i];
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}
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ans.push(s);
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}
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ans
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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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@ -208,12 +224,7 @@ function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
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* @return {number[]}
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*/
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var sumEvenAfterQueries = function (nums, queries) {
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let s = 0;
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for (const x of nums) {
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if (x % 2 === 0) {
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s += x;
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}
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}
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let s = nums.reduce((acc, cur) => acc + (cur % 2 === 0 ? cur : 0), 0);
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const ans = [];
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for (const [v, i] of queries) {
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if (nums[i] % 2 === 0) {
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@ -229,6 +240,31 @@ var sumEvenAfterQueries = function (nums, queries) {
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};
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```
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#### C#
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```cs
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public class Solution {
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public int[] SumEvenAfterQueries(int[] nums, int[][] queries) {
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int s = nums.Where(x => x % 2 == 0).Sum();
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int[] ans = new int[queries.Length];
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for (int j = 0; j < queries.Length; j++) {
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int v = queries[j][0], i = queries[j][1];
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if (nums[i] % 2 == 0) {
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s -= nums[i];
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}
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nums[i] += v;
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if (nums[i] % 2 == 0) {
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s += nums[i];
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}
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ans[j] = s;
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}
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return ans;
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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@ -60,7 +60,13 @@ After adding 2 to nums[3], the array is [-2,-1,3,6], and the sum of even values
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Simulation
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We use an integer variable $\textit{s}$ to record the sum of all even numbers in the array $\textit{nums}$. Initially, $\textit{s}$ is the sum of all even numbers in the array $\textit{nums}$.
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For each query $(v, i)$, we first check if $\textit{nums}[i]$ is even. If $\textit{nums}[i]$ is even, we subtract $\textit{nums}[i]$ from $\textit{s}$. Then, we add $v$ to $\textit{nums}[i]$. If $\textit{nums}[i]$ is even, we add $\textit{nums}[i]$ to $\textit{s}$, and then add $\textit{s}$ to the answer array.
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The time complexity is $O(n + m)$, where $n$ and $m$ are the lengths of the arrays $\textit{nums}$ and $\textit{queries}$, respectively. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.
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<!-- tabs:start -->
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@ -171,12 +177,7 @@ func sumEvenAfterQueries(nums []int, queries [][]int) (ans []int) {
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```ts
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function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
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let s = 0;
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for (const x of nums) {
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if (x % 2 === 0) {
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s += x;
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}
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}
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let s = nums.reduce((acc, x) => acc + (x % 2 === 0 ? x : 0), 0);
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const ans: number[] = [];
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for (const [v, i] of queries) {
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if (nums[i] % 2 === 0) {
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@ -192,6 +193,31 @@ function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn sum_even_after_queries(mut nums: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
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let mut s: i32 = nums.iter().filter(|&x| x % 2 == 0).sum();
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let mut ans = Vec::with_capacity(queries.len());
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for query in queries {
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let (v, i) = (query[0], query[1] as usize);
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if nums[i] % 2 == 0 {
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s -= nums[i];
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}
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nums[i] += v;
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if nums[i] % 2 == 0 {
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s += nums[i];
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}
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ans.push(s);
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}
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ans
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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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@ -201,12 +227,7 @@ function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
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* @return {number[]}
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*/
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var sumEvenAfterQueries = function (nums, queries) {
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let s = 0;
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for (const x of nums) {
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if (x % 2 === 0) {
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s += x;
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}
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}
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let s = nums.reduce((acc, cur) => acc + (cur % 2 === 0 ? cur : 0), 0);
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const ans = [];
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for (const [v, i] of queries) {
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if (nums[i] % 2 === 0) {
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@ -222,6 +243,31 @@ var sumEvenAfterQueries = function (nums, queries) {
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};
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```
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#### C#
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```cs
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public class Solution {
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public int[] SumEvenAfterQueries(int[] nums, int[][] queries) {
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int s = nums.Where(x => x % 2 == 0).Sum();
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int[] ans = new int[queries.Length];
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for (int j = 0; j < queries.Length; j++) {
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int v = queries[j][0], i = queries[j][1];
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if (nums[i] % 2 == 0) {
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s -= nums[i];
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}
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nums[i] += v;
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if (nums[i] % 2 == 0) {
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s += nums[i];
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}
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ans[j] = s;
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}
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return ans;
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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@ -0,0 +1,20 @@
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public class Solution {
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public int[] SumEvenAfterQueries(int[] nums, int[][] queries) {
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int s = nums.Where(x => x % 2 == 0).Sum();
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int[] ans = new int[queries.Length];
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for (int j = 0; j < queries.Length; j++) {
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int v = queries[j][0], i = queries[j][1];
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if (nums[i] % 2 == 0) {
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s -= nums[i];
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}
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nums[i] += v;
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if (nums[i] % 2 == 0) {
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s += nums[i];
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}
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ans[j] = s;
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}
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return ans;
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}
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}
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@ -4,12 +4,7 @@
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* @return {number[]}
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*/
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var sumEvenAfterQueries = function (nums, queries) {
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let s = 0;
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for (const x of nums) {
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if (x % 2 === 0) {
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s += x;
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}
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}
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let s = nums.reduce((acc, cur) => acc + (cur % 2 === 0 ? cur : 0), 0);
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const ans = [];
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for (const [v, i] of queries) {
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if (nums[i] % 2 === 0) {
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@ -0,0 +1,20 @@
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impl Solution {
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pub fn sum_even_after_queries(mut nums: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
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let mut s: i32 = nums.iter().filter(|&x| x % 2 == 0).sum();
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let mut ans = Vec::with_capacity(queries.len());
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for query in queries {
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let (v, i) = (query[0], query[1] as usize);
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if nums[i] % 2 == 0 {
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s -= nums[i];
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}
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nums[i] += v;
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if nums[i] % 2 == 0 {
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s += nums[i];
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}
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ans.push(s);
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}
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ans
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}
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}
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@ -1,10 +1,5 @@
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function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
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let s = 0;
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for (const x of nums) {
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if (x % 2 === 0) {
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s += x;
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}
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}
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let s = nums.reduce((acc, x) => acc + (x % 2 === 0 ? x : 0), 0);
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const ans: number[] = [];
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for (const [v, i] of queries) {
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if (nums[i] % 2 === 0) {
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