Leetcode 1005. Maximize Sum Of Array After K Negations

最新推荐文章于 2025-06-11 20:14:43 发布

原创最新推荐文章于 2025-06-11 20:14:43 发布 · 159 阅读

0 ·

CC 4.0 BY-SA版权

算法同时被 3 个专栏收录

475 篇文章

订阅专栏

leetcode

231 篇文章

订阅专栏

python3

120 篇文章

订阅专栏

本文探讨了如何通过选择特定元素并将其取反K次来最大化数组的总和。提供了三个示例，展示了如何通过这一过程使数组的总和达到最大。文章包括了一个Python类解决方案，其中包含了一个用于修改数组并返回最大可能总和的方法。

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

-----------------------------------------------------------------

非常简单，居然能wa一次，不应该啊。。。自己想一下，确保脑子里没有bug

class Solution:
    def largestSumAfterKNegations(self, A, K):
        sa = sorted(A)
        i = 0
        for i in range(K):
            if (sa[i] < 0):
                sa[i] = -sa[i]
            else:
                min_val = min(sa)
                res = sum(sa)
                delta = min_val
                for j in range(K - i):
                    res -= 2*delta
                    delta = -delta
                return res
        return sum(sa)
s = Solution()
print(s.largestSumAfterKNegations([4,2,3], 1))