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]]))