【leetcode】985. Sum of Even Numbers After Queries

Bertha 。 2022-01-07 11:25 156阅读 0赞

**题目如下：**

> 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`

**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

转载于:https://www.cnblogs.com/seyjs/p/10351178.html