【HackerRank】Find the Median(Partition找到数组中位数)

In the Quicksort challenges, you sorted an entire array. Sometimes, you just need specific information about a list of numbers, and doing a full sort would be unnecessary. Can you figure out a way to use your partition code to find the median in an array?

Challenge 
Given a list of numbers, can you find the median?

Input Format 
There will be two lines of input:

• n - the size of the array
• ar - n numbers that makes up the array

Output Format 
Output one integer, the median.

Constraints 
1<= n <= 1000001 
-10000 <= x <= 10000 , x ∈ ar 
There will be an odd number of elements.

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题解：久闻Partition函数可以用来找到乱序数组的中位数，今天终于实现了一把。

设置一个变量need为数组长度的一半，另一个变量hasFound为当前比找到的比中位数小的数的个数，当hasFound=need的时候，我们就找到了中位数。

在每次Partition后，看比pivot小的那部分数组有多少个元素，如果hasFound加上这部分元素正好等于need，那么pivot就是所求的中位数。如果hasFound加上这部分元素大于need，说明比pivot小的这部分数组需要继续划分，递归的调用Partition；如果hasFound加上这部分元素小于need，那么就把这部分元素个数加到hasFound上，并且继续划分比pivot大的那部分数组。

例如数组：3 1 4 5 2，找中位数的过程如下图所示：

代码如下：

 1 import java.util.*;

 2 

 3 public class Solution {

 4     private static int need = 0;

 5     private static int hasFound = 0;

 6     

 7     private static void swap(int[] ar,int i,int j){

 8         int temp = ar[i];

 9         ar[i]= ar[j];

10         ar[j]=temp;

11     }

12     private static int Partition(int[] ar,int start,int end){

13         int pivot = ar[end];

14         int i = start;

15         int j = start;

16         while(i < end){

17             if(ar[i] >= pivot)

18                 i++;

19             else if(ar[i] < pivot){

20                 swap(ar,i,j);

21                 i++;

22                 j++;

23             }

24         }

25         swap(ar, j, end);

26         if(hasFound+j-start+1==need)

27             return pivot;

28         else if(hasFound+j-start+1<need){

29             hasFound += j-start+1;

30             return Partition(ar, j+1, end);

31         }

32         else {

33             return Partition(ar, start, j-1);

34         }

35         

36     }

37     

38     public static void main(String[] args) {

39         Scanner in = new Scanner(System.in);

40         int n = in.nextInt();

41         need = n%2==0?n/2:n/2+1;

42         int[] ar = new int[n];

43         

44         for(int i = 0;i < n;i ++)

45             ar[i] = in.nextInt();

46         System.out.println(Partition(ar, 0, n-1));

47         

48     }

49 }

以上代码就很容易扩展到找寻数组中第k小的数了，只要改变need变量的值即可。