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:

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].

根据题意，想找到所有可能组合中最小数和的最大值。按照题目给出的例子，结合网上查找的数学证明，要按从小到大的顺序两两组合，再取最小的数则可使其和最大。所以我们先对数组从小到大排序 ，然后跳位相加即为所求。（感觉和two-pointer也没啥关系呀-.-）

 

 1 class Solution {
 2 public:
 3 int arrayPairSum(vector<int>&nums) {
 4     int result=0;
 5     //先把数组按从小到大的顺序进行排序
 6     sort(nums.begin(), nums.end());
 7     for (int i = 0; i < nums.size(); i=i+2) {
 8         result += nums[i];
 9     }
10     return result;
11 }
12 };