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: 4Explanation: n is 2, and the maximum sum of pairs is 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].

这个问题，凭直觉就知道是将数字从小到大排序，然后每两组取最小的相加。于是我写了个冒泡，结果run超时了，好吧，逼我写了个快排。

public class Array_Partition_I_561 {public void quickSort(int[] list,int left,int right){if (left < right) {int pos = partition(list, left, right);quickSort(list, left, pos - 1);quickSort(list, pos + 1, right);}}public int partition(int[] list,int low,int high){int pos=list[low];while(low<high){while(list[high]>=pos&&low<high){high--;}if(low<high){list[low]=list[high];low++;}while(list[low]<=pos&&low<high){low++;}if(low<high){list[high]=list[low];high--;}}list[low]=pos;return low;}public int arrayPairSum(int[] nums) {        quickSort(nums, 0, nums.length-1);        int sum=0;        for(int i=0;i<nums.length/2;i++){        sum+=nums[i*2];        }        return sum;    }public static void main(String[] args) {Array_Partition_I_561 a=new Array_Partition_I_561();int[] nums=new int[]{1,4,3,2};    System.out.println(a.arrayPairSum(nums));}}

终于AC了，我看了看其他大神，发现很多人直接

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

喂喂喂。。虽然这样可以。。。。-_-||

有大神给出了排序后就能得到每对最小值之和最大的证明，very nice！

①.假设在每对(ai,bi)中，bi>=ai（顺序不影响min，所以调换顺序没有关系）

②.令 Sm = min(a1, b1) + min(a2, b2) + ... + min(an, bn) ，那么该题所要求的就是最大的Sm。由①可得，Sm = a1 + a2 + ... + an.

③.令Sa = a1 + b1 + a2 + b2 + ... + an + bn. Sa是一个常量。

④.令di = |ai - bi|. 由①可得，di = bi - ai. 令Sd = d1 + d2 + ... + dn.

⑤.所以Sa = a1 + a1 + d1 + a2 + a2 + d2 + ... + an + an + di = 2Sm + Sd => Sm = (Sa - Sd) / 2. 为了得到最大的Sm，考虑到Sa是一个常量，所以我们期望Sd最小。

⑥所以这个问题转化为另一个问题：找到一个数组中的pairs (ai,bi)，使得|ai-bi|（ai和bi之间的距离）之和最小。显而易见的是，相邻元素距离之和是最小的。直观的可见下图。