[leetcode] 985. Sum of Even Numbers After Queries

[leetcode] 985. Sum of Even Numbers After queries

Description

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.

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] <= 100001 <= queries.length <= 10000-10000 <= queries[i][0] <= 100000 <= queries[i][1] < A.length

分析

题目的意思是：维护一个数组，每个query过来的时候更新一下数组，然后把数组中偶数的和返回。我一开始用暴力破解的方法，很不幸超时了，最后试了一下维护ans，ans代表数组中所有偶数的和，然后query过来的时候，我就动态的更新ans，写得没有标准答案简洁，但我认为还不错了，至少时间复杂度是跟标准答案一样了哈哈

代码

class Solution: def sumEvenAfterQueries(self, A: List[int], queries: List[List[int]]) -> List[int]: res=[] ans=sum(val for val in A if(val%2==0)) for query in queries: val=query[0] index=query[1] if((A[index]%2)==(val%2)): if(val%2==1): ans+=A[index]+val else: ans+=val A[index]+=val res.append(ans) else: if(A[index]%2==0): ans-=A[index] res.append(ans) A[index]+=val return res

参考文献

​​[LeetCode] Approach 1: Maintain Array Sum​​

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