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.
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
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