2019-03-29-sum of even numbers after queries

题目，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.

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.

解析

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var sumEvenAfterQueries = (A, queries) => {
  let evenSum = A.reduce((acc, cur) => acc + (cur % 2 ? 0 : cur), 0); //数组里面偶数求和
  return queries.map(([val, idx]) => {
    if(A[idx] % 2 === 0) evenSum -= A[idx]
    A[idx] += val
    if(A[idx] % 2 === 0) evenSum += A[idx]
    return evenSum
  });
}