题目如下:
We have an array
A
of integers, and an arrayqueries
of queries.For the
i
-th queryval = queries[i][0], index = queries[i][1]
, we add val toA[index]
. Then, the answer to thei
-th query is the sum of the even values ofA
.(Here, the given
index = queries[i][1]
is a 0-based index, and each query permanently modifies the arrayA
.)Return the answer to all queries. Your
answer
array should haveanswer[i]
as the answer to thei
-th query.Example 1:
Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]] Output: [8,6,2,4] Explanation: At the beginning, the array is [1,2,3,4]. After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8. After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6. After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2. After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.Note:
1 <= A.length <= 10000
-10000 <= A[i] <= 10000
1 <= queries.length <= 10000
-10000 <= queries[i][0] <= 10000
0 <= queries[i][1] < A.length
解题思路:题目很简单,有一点需要注意,就是每次执行query后,不需要遍历整个数组求出所有偶数的值。只需要记录上一次偶数的和,执行query前,如果该值是偶数,用上一次的和减去该值;执行query后,如果新值是偶数,用上一次的和加上这个新值,就可以得到这次query执行后的偶数总和。
代码如下:
class Solution(object): def sumEvenAfterQueries(self, A, queries): """ :type A: List[int] :type queries: List[List[int]] :rtype: List[int] """ res = [] count = None for val,inx in queries: before = A[inx] A[inx] += val after = A[inx] if count == None: count = sum(filter(lambda x: x % 2 == 0,A)) else: if before % 2 == 0: count -= before if after % 2 == 0: count += after res.append(count) return res
原文地址:https://www.cnblogs.com/seyjs/p/10351178.html
时间: 2024-10-04 04:25:37