561.arrayPairSum

最新推荐文章于 2024-08-06 23:36:20 发布

Lxiaohuli

最新推荐文章于 2024-08-06 23:36:20 发布

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本文介绍了一种算法，用于将给定的2n个整数分成n对，使得每对中较小数之和尽可能大。通过排序和选择每对中的第一个元素求和，实现了高效解决方案。

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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).

#include <iostream>
#include <vector>

#include<algorithm>
using namespace std;

class Solution {
public:
	int arrayPairSum(vector<int>& nums) {
		int sum = 0;
 
		sort(nums.begin(),nums.end());  //排序
		for (int i = 0; i < nums.size() - 1; i=i+2)//求和
		{
			sum = sum + nums[i];
		}

		return sum;
	}
};

sort()比我写的快多了