leetcode-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:
n is a positive integer, which is in the range of [1, 10000].
All the integers in the array will be in the range of [-10000, 10000].

思路：
把元素分两个一组，求每对元素里面最小的和，使这个和最大。
其实按顺序排序每组选个最小的就可以。
因为如果不按顺序，比如把一个大的和一个很小的放一起，你这时候选那个小的，大数值就有点浪费了。

代码：

class Solution {public:    int arrayPairSum(vector<int>& nums) {        sort(nums.begin(),nums.end());        int size=nums.size();        int sum=0;        for(int i=0;i<size;i+=2){//每对里最小和下一对最小间隔是2            sum+=nums[i];        }        return sum;    }};