本文主要是介绍LeetCode1005. Maximize Sum Of Array After K Negations，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
文章目录
- 一、题目
- 二、题解
一、题目
Given an integer array nums and an integer k, modify the array in the following way:
choose an index i and replace nums[i] with -nums[i].
You should apply this process exactly k times. You 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: nums = [4,2,3], k = 1
Output: 5
Explanation: Choose index 1 and nums becomes [4,-2,3].
Example 2:
Input: nums = [3,-1,0,2], k = 3
Output: 6
Explanation: Choose indices (1, 2, 2) and nums becomes [3,1,0,2].
Example 3:
Input: nums = [2,-3,-1,5,-4], k = 2
Output: 13
Explanation: Choose indices (1, 4) and nums becomes [2,3,-1,5,4].
Constraints:
1 <= nums.length <= 104
-100 <= nums[i] <= 100
1 <= k <= 104
二、题解
class Solution {
public:static bool cmp(int a,int b){return abs(a) > abs(b);}int largestSumAfterKNegations(vector<int>& nums, int k) {int n = nums.size();sort(nums.begin(),nums.end(),cmp);for(int i = 0;i < n;i++){if(nums[i] < 0 && k > 0){nums[i] *= -1;k--;}}if(k % 2 == 1) nums[n-1] *= -1;int res = 0;for(int i = 0;i < n;i++){res += nums[i];}return res;}
};
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