[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. 1 <= A.length <= 10000
  2. -10000 <= A[i] <= 10000
  3. 1 <= queries.length <= 10000
  4. -10000 <= queries[i][0] <= 10000
  5. 0 <= 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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