561. Array Partition I

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 the sum of min(ai, bi) for all i from 1 to n as large as possible.

Example

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).

Example 2:

Input: [5,6]
Output: 5
Explanation: n is 1, and the maximum sum of pairs is 5 = min(5, 6) .

Notice

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].

1. class Solution {  
2. public:  
3.     int arrayPairSum(vector<int>& nums) {  
4.         sort(nums.begin(), nums.end());  
5.           
6.         int sum = 0;  
7.           
8.         for (int i = 0; i < nums.size(); i += 2) {  
9.             sum += nums[i];  
10.         }  
11.           
12.         return sum;  
13.     }  
14. };  

Note:

The summation value is the largest. Add all the smallest values ​​in a set of pairs.