Leetcode 985. Sum of Even Numbers After Queries

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/M_sdn/article/details/86760532

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

public int[] sumEvenAfterQueries(int[] A, int[][] queries) {
        int[] ans = new int[A.length];
        for (int i = 0; i < queries.length; i++) {
            int[] query = queries[i];
            int add = query[0];
            int index = query[1];
            A[index] += add;
            int sum = 0;
            for (int j = 0; j < A.length; j++) {
                if (A[j] % 2 == 0) {
                    sum += A[j];
                }
            }
            ans[i] = sum;
        }
        return ans;
    }