985. Sum of Even Numbers After Queries

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.

有两个数组A和queries，queries是个二维数组，第一个元素是值，第二个元素是索引，循环queries数组，把queries的每个值加到A数组对应索引位置上。求出每一次循环后A数组中的偶数和。

思路：先求出A数组中的偶数和。再循环queries数组，共四种情况：加运算之前是偶数，之后是奇数：总和-旧值；偶数-》偶数：总和+新值；奇数-》偶数：总和+旧值+新值；奇数-》奇数：总和不变。

class Solution {
    public int[] sumEvenAfterQueries(int[] A, int[][] queries) {
        int[] result = new int[queries.length];
        int evenSum = 0;
        for (int i1 : A) {
            if (i1 % 2 == 0) {
                evenSum += i1;
            }
        }
        for (int i = 0; i < queries.length; i++) {
            int index = queries[i][1];
            int val = queries[i][0];
            int old = A[index];

            A[index] = A[index] + val;
            if (old % 2 == 0 && A[index] % 2 != 0) {
                result[i] = evenSum - old;
                evenSum = result[i];
                continue;
            }
            if (old % 2 == 0 && A[index] % 2 == 0) {
                result[i] = evenSum + val;
                evenSum = result[i];
                continue;
            }
            if (old % 2 != 0 && A[index] % 2 == 0) {
                result[i] = evenSum + old + val;
                evenSum = result[i];
                continue;
            }
            if (old % 2 != 0 && A[index] % 2 != 0) {
                result[i] = evenSum;
                evenSum = result[i];
            }
        }
        return result;
    }
}