[Array] 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 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].

思路：将整个数组进行排序，然后每个数对的最小值就是左边的值，在数组中的显示就是隔一个数字，因此，排序后，以2为步进求和。

代码：

int arraypairsum(vector<int>& nums)
{
sort(nums.begin(), nums.end());
int sum = 0;
for(int i = 0; i < nums.size(); i = i + 2)
sum = sum + nums[i];
return sum;  
}