[leetcode]561. Array Partition I-优快云博客

本文介绍了一种针对2n个整数进行最优配对求和的问题解决方法。任务是将这些整数分成n组，每组两个数，并使得所有组中小数之和尽可能大。文章提供了具体的算法实现思路及Java代码示例。

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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;
	}
}