0985. Sum of Even Numbers After Queries

985. Sum of Even Numbers After Queries #

题目 #

We have an array A of integers, and an array queries of queries.

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.

(Here, the given index = queries[i][1] is a 0-based index, and each query permanently modifies the array A.)

Return the answer to all queries. Your answer array should have answer[i] as the answer to the i-th query.

Example 1:

Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
Output: [8,6,2,4]
Explanation: 
At the beginning, the array is [1,2,3,4].
After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.

Note:

  1. 1 <= A.length <= 10000
  2. -10000 <= A[i] <= 10000
  3. 1 <= queries.length <= 10000
  4. -10000 <= queries[i][0] <= 10000
  5. 0 <= queries[i][1] < A.length

题目大意 #

给出一个整数数组 A 和一个查询数组 queries。

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。

解题思路 #

  • 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。
  • 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。

代码 #


package leetcode

func sumEvenAfterQueries(A []int, queries [][]int) []int {
	cur, res := 0, []int{}
	for _, v := range A {
		if v%2 == 0 {
			cur += v
		}
	}
	for _, q := range queries {
		if A[q[1]]%2 == 0 {
			cur -= A[q[1]]
		}
		A[q[1]] += q[0]
		if A[q[1]]%2 == 0 {
			cur += A[q[1]]
		}
		res = append(res, cur)
	}
	return res
}


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