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.

给定一个整数数组 A，我们只能用以下方法修改该数组：我们选择某个个索引 i 并将 A[i] 替换为 -A[i]，然后总共重复这个过程 K 次。（我们可以多次选择同一个索引 i。）

以这种方式修改数组后，返回数组可能的最大和。

题目链接：https://leetcode.com/problems/maximize-sum-of-array-after-k-negations/

个人主页：http://redtongue.cn or https://redtongue.github.io/

Difficulty: easy

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

• 翻转，应尽可能的翻转负数，同时翻转两次是可以抵消的，故分为如下三种情况；
• 1、K小于负数的个数，翻转最小的K个负数；
• 2、K大于负数的个数，且K-负数的个数等于偶数，翻转所有负数；
• 3、K大于负数的个数，且K-负数的个数等于奇数，翻转所有负数；
• 对于1，2返回翻转之后数组的和，对于3，返回翻转之后数组的和减去数组中最小数字的两倍。

参考代码

class Solution(object):
def largestSumAfterKNegations(self, A, K):
    A.sort()
    judge=True
    for i in range(K):
        if(A[i]<0):
            A[i]=-A[i]
        else:
            if((K-i)%2==0):
                judge=True
            else:
                judge=False
            break
    if(judge):
        return sum(A)
    else:
        return sum(A)-2*min(A)