LeetCode 561. Array Partition I

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].

问题分析

给定一个长度为2n的数组，将这个数组分为n个(ai，bi)，其中ai

sum(ai) + sum（bi） = sun(all)==> sum（ai）+sum(bi-ai)+sum（ai）= sum（all）==> 2sum（ai）+sum(bi-ai) = sum（all）==>sum（all）是常数，求sum(ai)最大，变为求sum(bi-ai)的和最小。==> 相邻大小的两个值的差值的和最小。

代码实现

 public int arrayPairSum(int[] nums) {        if (nums == null || nums.length == 0) {            return 0;        }        Arrays.sort(nums);        int sum = 0;        for (int i = 0; i < nums.length; i += 2) {            sum += nums[i];        }        return sum;    }