561. Array Partition I

Given an array of 2n integers, your task is to group these integers inton 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: 4Explanation: 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].

分析：

       题目比较简单，对2n个数进行两两分组，取出每组中的最小数要求组成的和最大。这里先对数组进行

从小到大排列，然后取出数组中的奇数项进行相加即可得到答案。

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