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].

题目意思：从2n数组两两取数，形成n对（ai，bj），从这n对整数中选出每对中最小的数，相加的和最大。

分析：将2n数组排序，依次去下标为0,2,4，i+2...的数。

c++代码：

#include <algorithm>
class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        int sum=0;
        vector<int>::iterator it;
        for(it=nums.begin();it!=nums.end();it+=2)
            sum+=*it;
        return sum;
    }
};