LeetCode1005. Maximize Sum Of Array After K Negations

文章目录

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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