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].

这个数学问题感觉比较简单？要使总和最大，就要把最小的数和次小的数分在同一组，不然它们都会被用来做加法
那思路就是做一个由小到大的排序，然后从第一个数字开始，隔一个取一个数字相加就是结果了

python内置函数多，可以直接排序、求和

class Solution(object):    def arrayPairSum(self, nums):        """        :type nums: List[int]        :rtype: int        """        return sum(sorted(nums)[::2])

C++同理，不过求和的时候用for循环

class Solution {public:    int arrayPairSum(vector<int>& nums) {        int sum = 0;        int i = 0;        int size = nums.size();        sort(nums.begin(), nums.end());        for(i=0; i<size-1; i+=2){            sum += nums[i];        }               return sum;    }};