561. Array Partition I 数组划分1

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.
给定含2n个元素的数组，将其分为n个二元组，使得每个二元组内较小的值的总和最小.

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
排序数组后将相邻元素划分为一组，取其中较小的，实际上就是偶数位的元素。
不过本题只要求算法返回最小的和，可理解位2n个元素中取n个使得总和最小，可直接排序后输出前n个和即可。

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