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
