LeetCode1005. Maximize Sum Of Array After K Negations

本文介绍: 【代码】Lee tCod e1005. Maxi m ize Su m Of Ar r a y Aft er K Neg a t ion s。

文章 目录

- 一、题目
- 二、题解

一、题目

Gi ven a n integer array nums and a n integer k, modify the array in the f ol low in g w a y:

ch o ose a n index i and replace nums[i] wi th –nums[i].
You sh ould apply th i s process exa ctly k times. You may ch o ose the same i ndex i multiple times.

Return the largest p oss ible sum of t he array after modify in g i t in t hi s w a y.

Ex ample 1:

Inpu t: nums = [4,2,3], k = 1
Out put: 5
Ex plan a t ion: Choose index 1 and nums beco mes [4,-2,3].
Example 2:

Inpu t: nums = [3,-1,0,2], k = 3
Output: 6
Explanation: Choose in dic es (1, 2, 2) and nums beco mes [3,1,0,2].
Example 3:

Input: nums = [2,-3,-1,5,-4], k = 2
Output: 13
Explanation: Choose in dic es (1, 4) and nums becomes [2,3,-1,5,4].

Con s train ts:

1 <= nums.length <= 104
-100 <= num s[i] <= 100
1 <= k <= 104

二、题解

class Solution {
public:
    static bool cmp(int a,int b){
        return abs(a) &gt; abs(b);
    }
    int largestSumAfterKNegations(vector<int&gt;&amp; nums, int k) {
        int n = nums.size();
        sort(nums.begin(),nums.end(),cmp);
        for(int i = 0;i < n;i++){
            if(nums[i] < 0 &amp;&amp; k &gt; 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;
    }
};