[leetcode]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.

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].

分析：

2n个数字分成n组，每组2个数字，取每组中较小的数字，求和。要求是得到的和最大，方法返回值即为这个最大和。

设2n个数字升序排序后为：a0，a1，a2，a3，….，a(2n-1)。  猜想一下容易发现，最大和即为 (a0+a2+a4+….+a(2n-2))

/** * @author binkang * @date May 6, 2017 */ public class ArrayPartition1 { public int arrayPairSum(int[] nums) { Arrays.sort(nums); int sum = 0; for(int i=0;i<nums.length; i = i+2) { sum += getMin(nums[i], nums[i+1]); } return sum; } public int getMin(int a, int b) { return a < b ? a : b; } }