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), ..., (a_n, b_n) which makes sum of min(a_i, b_i) 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].

解题思路：先将该数组排序，然后取每组的最小值之和，即为所求。

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        if (nums.size == 0)

           return 0;

        sort (nums, nums + nums.size);//排序

       int sum = 0;

       for（int i = 0; i < nums.size; i += 2）{

               sum += nums[i];

         }

         return sum;

};