Array Partition I

1. 问题描述

   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].

   问题比较简单，其实为给定一个int类型的数组，然后对数组进行两两分组，但是分组之后需要确保两两分组之后的min值的sum为最大，退而求其次，求两两分组后max值sum最大即为数组中排序然后将后面大的数子进行相加，那第二大无非就是排序后，两两分组即可，可以将损失降到最低

2.代码

 public int arrayPairSum(int[] nums) {        int sum = 0;        Arrays.sort(nums);        for (int i =0;i<nums.length;i++){            if ((i & 1) == 0){                sum += nums[i];            }        }        return sum;    }