Leetcode-Easy 985. Sum of Even N

题目描述

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.

思路

新建一个列表，保留更新后的偶数值，将奇数值设为0，然后求和

代码实现

class Solution:
    def sumEvenAfterQueries(self, A: 'List[int]', queries: 'List[List[int]]') -> 'List[int]':
        new_A=[None]*len(A)
        even_index=[a if a%2==0 else 0 for a in A]
        for i in range(len(A)):
            val = queries[i][0]
            index = queries[i][1]
            A[index]+=val
            if A[index]%2==0:
                even_index[index]=A[index]
            else:
                even_index[index]=0
            new_A[i]=sum(even_index)
        return new_A