[leetcode]561. Array Partition I

题目链接

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:

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].

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概要
给定一个包含 2n 个整数的数组，任务是将其以 2 个为一组进行分组，分成 n 组，(a1, b1), (a2, b2), … ,(an, bn), 并使得

∑i=1nmin(ai,bi)

为最大值.

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思路
首先对于给定的 n，总和 St 是确定的，即
St = (a1 + b1) + (a2 + b2) + … + (an + bn)
题目要求的最大的和为
Sm = a1 + a2 + … + an
其中有
Sd = (b1 - a1) + (b2 - a1) + … + (bn - an)
可得
Sm = (St - Sd) / 2
则当 Sd 为最小时，Sm有最大值
当 a1, b1, …, an, bn 按大小排列组合时，Sd 有最小值

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代码实现

public class Solution {    public int arrayPairSum(int[] nums) {        Arrays.sort(nums);        int length = nums.length / 2 - 1;        int sum = 0;        while (length >= 0) {            sum += nums[2 * length--];        }        return sum;    }}