[LC][array] Add to List 561. Array Partition I

一、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: 4Explanation: 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].

二、我的解法

因为要求min()的sum最大。min取决于最小的那个数字，所以如果一个小的数字拉一个大的数字垫背的话，就会损失很多。最好的方法是让两个相近的数字组合在一起，损失的小。

于是乎，先排序，然后sum的时候步长为2加起来就行了。

class Solution {    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 = i + 2){            sum += nums[i];        }                return sum;    }}

三、淫奇技巧

已经是目前最优了。但是大神给出了证明……

1. Assume in each pair i, bi >= ai.
2. Denote Sm = min(a1, b1) + min(a2, b2) + ... + min(an, bn). The biggest Sm is the answer of this problem. Given 1, Sm = a1 + a2 + ... + an.
3. Denote Sa = a1 + b1 + a2 + b2 + ... + an + bn. Sa is constant for a given input.
4. Denote di = |ai - bi|. Given 1, di = bi - ai. Denote Sd = d1 + d2 + ... + dn.
5. So Sa = a1 + a1 + d1 + a2 + a2 + d2 + ... + an + an + di = 2Sm + Sd => Sm = (Sa - Sd) / 2. To get the max Sm, given Sa is constant, we need to make Sd as small as possible.
6. So this problem becomes finding pairs in an array that makes sum of di (distance between ai and bi) as small as possible. Apparently, sum of these distances of adjacent elements is the smallest. If that's not intuitive enough, see attached picture. Case 1 has the smallest Sd

四、知识点&注意事项

Arrays.sort()这个东西之前不知道。。

Array有关的操作：

http://www.codeceo.com/article/10-java-array-method.html

http://www.jianshu.com/p/bef8c1b93555

五、扩展问题

如果不给定2n会怎样？要注意单数的情况，从1开始取。

如果要求sum最小？排序后加最小的那半。

六、碎碎念

这个题目解得挺快。。。明天继续加油