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).

 给定一个长度为2n(偶数)的数组，分成n个小组，返回每组中较小值的和sum，使sum尽量大

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].

思路：

先排序，将相邻两个数分为一组，每组较小数都在左边，求和即可

 1     public int arrayPairSum(int[] nums) {
 2         Arrays.sort(nums);
 3         
 4         int sum = 0;
 5         for (int i=0;i<nums.length;i+=2)
 6         {
 7             sum += nums[i];
 8         }
 9         return sum;
10     }

561. Array Partition I

原文：http://www.cnblogs.com/wzj4858/p/7670011.html