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].

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/array-partition-i
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

感觉有的人为了提速，真的是啥都敢干，，，，

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
         if(nums.empty())return 0;
        sort(nums.begin(),nums.end());
        int t=0,sum=0;
        while(t<nums.size()){
            sum+=nums[t];
            t+=2;
        }return sum;
    }
};