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]

   class Solution {    public int arrayPairSum(int[] nums) {        if(nums==null||nums.length<=1)return 0;        Arrays.sort(nums);        int count=0;        for(int i=0;i<nums.length;i=i+2){            count=count+nums[i];        }        return count;    }}

   数组元素两两组合，计算所有组合后的两个数的最小值之和，考虑如何使得这个和最大；最简单的方式就是将数组重新排序，然后计算奇数位置的数值之和即可