[leetcode] 1005. Maximize Sum Of Array After K Negations

Description

Given an array A of integers, we must modify the array in the following way: we choose an i and replace A[i] with -A[i], and we repeat this process K times in total. (We may choose the same index i multiple times.)

Return the largest possible sum of the array after modifying it in this way.

Example 1:

Input: A = [4,2,3], K = 1
Output: 5
Explanation: Choose indices (1,) and A becomes [4,-2,3].

Example 2:

Input: A = [3,-1,0,2], K = 3
Output: 6
Explanation: Choose indices (1, 2, 2) and A becomes [3,1,0,2].

Example 3:

Input: A = [2,-3,-1,5,-4], K = 2
Output: 13
Explanation: Choose indices (1, 4) and A becomes [2,3,-1,5,4].

Note:

• 1 <= A.length <= 10000
• 1 <= K <= 10000
• -100 <= A[i] <= 100

分析

题目的意思是：给定数组A，替换其中的数字变成相反数，同一个数字可以替换无限次。

• 这道题需要用到最小堆，每次把最小的数取出来，然后变成相反数，这样得到的数组的求和就是最大的哈。

代码

class Solution:
    def largestSumAfterKNegations(self, A: List[int], K: int) -> int:
        res=sum(A)
        heapq.heapify(A)
        while(K>0):
            cur=heapq.heappop(A)
            heapq.heappush(A,-cur)
            K-=1
            res=res-cur+(-cur)
        return res

参考文献

【LeetCode】1005. Maximize Sum Of Array After K Negations 解题报告（Python）