LeetCode 561. Array Partition I

Posted A-Little-Nut

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了LeetCode 561. Array Partition I相关的知识,希望对你有一定的参考价值。

Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), …, (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.

Example 1:

Input: [1,4,3,2]
Output: 4
Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).

Note:

  • n is a positive integer, which is in the range of [1, 10000].
  • All the integers in the array will be in the range of [-10000, 10000].
class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        int sum=0;
        sort(nums.begin(),nums.end());
        for(int i=nums.size()-2;i>=0;i-=2)
            sum+=nums[i];
        return sum;
            
    }
};

以上是关于LeetCode 561. Array Partition I的主要内容,如果未能解决你的问题,请参考以下文章

leetcode-561(Array Partition I)

leetcode-561(Array Partition I)

leetcode-561-Array Partition I

[LeetCode] 561. Array Partition I

LeetCode 561. Array Partition I

561. Array Partition I - LeetCode