985. 查询后的偶数和

代码一：暴力超时。

class Solution(object):
    # 暴力超时
    def sumEvenAfterQueries(self, A, queries):
        """
        :type A: List[int]
        :type queries: List[List[int]]
        :rtype: List[int]
        """
        ans = []
        for i in range(len(A)):
            A[queries[i][1]] += queries[i][0]
            temp = 0
            for j in range(len(A)):
                if A[j] % 2 == 0:
                    temp += A[j]
            ans.append(temp)
        return ans


if __name__ == '__main__':
    solution = Solution()
    print(solution.sumEvenAfterQueries(A=[1, 2, 3, 4], queries=[[1, 0], [-3, 1], [-4, 0], [2, 3]]))

代码二：

 1 class Solution(object):
 2     def sumEvenAfterQueries(self, A, queries):
 3         """
 4         :type A: List[int]
 5         :type queries: List[List[int]]
 6         :rtype: List[int]
 7         """
 8         # 返回值
 9         ans = []
10         # 统计原list中偶数元素的和
11         temp = 0
12         for i in range(len(A)):
13             if A[i] % 2 == 0:
14                 temp += A[i]
15         for j in range(len(A)):
16             val = queries[j][0]
17             index = queries[j][1]
18             # 若将要改变的元素是偶数，则要从和中减掉
19             if A[index] % 2 == 0:
20                 temp -= A[index]
21             # 改变元素
22             A[index] += val
23             # 若改变以后是偶数，则要加到和中
24             if A[index] % 2 == 0:
25                 temp += A[index]
26             ans.append(temp)
27         return ans
28 
29 
30 if __name__ == '__main__':
31     solution = Solution()
32     print(solution.sumEvenAfterQueries(A=[1, 2, 3, 4], queries=[[1, 0], [-3, 1], [-4, 0], [2, 3]]))