561. Array Partition I

一、问题：

 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: 4Explanation: 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].

二、解题：

class Solution {    /*      数组进行排序，偶数位的都是最小中可取的最大的数。      比如 [-1,1,2,3,4,5] max=-1+2+4=5;    */public:    int arrayPairSum(vector<int>& nums)    {        int n = nums.size();        int max = 0;        sort(nums.begin(), nums.end());        for (int i = 0; i < n - 1; i = i + 2)        {            max += nums[i];        }        return max;    }};