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

 

Note:

1. n is a positive integer, which is in the range of [1, 10000].
2. All the integers in the array will be in the range of [-10000, 10000].

思路：

需要注意的是找出配对后最小值的和的最大值，其实就是要求最小值要与另一个数的差值最小，这样最小值才有机会获得更大的值。这样，让每个配对的数中较大值取为较小值的后一个数。

class Solution {public:    int arrayPairSum(vector
     
      & nums) {        int res = 0;        sort(nums.begin(), nums.end());        for (int i = 0; i < nums.size(); i++){            if (i % 2 == 0){                res += nums[i];            }        }        return res;    }};