LeetCode 561. Array Partition I

题目：
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.
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].

思路：
就是给2n个数分组，两两一组，使用所有组中小的那个数加起来和最小。
那么就是先将所有数排序，然后就是以排序后的值2个一组分组，然后取的就是两个中前一个的索引对应的值。

代码：

class Solution {public:    int arrayPairSum(vector<int>& nums) {        vector<int>::size_type nsize=nums.size();//计算数组大小        if(nsize==0){//如果大小为0，直接返回0            return 0;        }        sort(nums.begin(),nums.end());//排序        int sum=0;        for(int i=0;i<nsize;i+=2){//计算两两一组中，数小的那个，和即为结果            sum+=nums[i];        }        return sum;    }};

输出结果： 106ms