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68 lines
2.4 KiB
Markdown
68 lines
2.4 KiB
Markdown
# [985. Sum of Even Numbers After Queries](https://leetcode.com/problems/sum-of-even-numbers-after-queries/)
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## 题目
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We have an array `A` of integers, and an array `queries` of queries.
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For the `i`-th query `val = queries[i][0], index = queries[i][1]`, we add val to `A[index]`. Then, the answer to the `i`-th query is the sum of the even values of `A`.
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*(Here, the given `index = queries[i][1]` is a 0-based index, and each query permanently modifies the array `A`.)*
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Return the answer to all queries. Your `answer` array should have `answer[i]` as the answer to the `i`-th query.
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**Example 1**:
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```
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Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
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Output: [8,6,2,4]
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Explanation:
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At the beginning, the array is [1,2,3,4].
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After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
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After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
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After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
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After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.
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```
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**Note**:
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1. `1 <= A.length <= 10000`
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2. `-10000 <= A[i] <= 10000`
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3. `1 <= queries.length <= 10000`
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4. `-10000 <= queries[i][0] <= 10000`
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5. `0 <= queries[i][1] < A.length`
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## 题目大意
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给出一个整数数组 A 和一个查询数组 queries。
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对于第 i 次查询，有 val = queries[i][0], index = queries[i][1]，我们会把 val 加到 A[index] 上。然后，第 i 次查询的答案是 A 中偶数值的和。（此处给定的 index = queries[i][1] 是从 0 开始的索引，每次查询都会永久修改数组 A。）返回所有查询的答案。你的答案应当以数组 answer 给出，answer[i] 为第 i 次查询的答案。
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## 解题思路
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- 给出一个数组 A 和 query 数组。要求每次 query 操作都改变数组 A 中的元素值，并计算此次操作结束数组 A 中偶数值之和。
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- 简单题，先计算 A 中所有偶数之和。再每次 query 操作的时候，动态维护这个偶数之和即可。
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## 代码
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```go
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func sumEvenAfterQueries(A []int, queries [][]int) []int {
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cur, res := 0, []int{}
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for _, v := range A {
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if v%2 == 0 {
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cur += v
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}
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}
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for _, q := range queries {
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if A[q[1]]%2 == 0 {
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cur -= A[q[1]]
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}
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A[q[1]] += q[0]
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if A[q[1]]%2 == 0 {
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cur += A[q[1]]
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}
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res = append(res, cur)
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}
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return res
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}
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``` |