【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：

 import java.util.*;

 public class Solution {
     private static int need = 0;
     private static int hasFound = 0;

     private static void swap(int[] ar,int i,int j){
         int temp = ar[i];
         ar[i]= ar[j];
         ar[j]=temp;
     }
     private static int Partition(int[] ar,int start,int end){
         int pivot = ar[end];
         int i = start;
         int j = start;
         while(i < end){
             if(ar[i] >= pivot)
                 i++;
             else if(ar[i] < pivot){
                 swap(ar,i,j);
                 i++;
                 j++;
             }
         }
         swap(ar, j, end);
         if(hasFound+j-start+1==need)
             return pivot;
         else if(hasFound+j-start+1<need){
             hasFound += j-start+1;
             return Partition(ar, j+1, end);
         }
         else {
             return Partition(ar, start, j-1);
         }

     }

     public static void main(String[] args) {
         Scanner in = new Scanner(System.in);
         int n = in.nextInt();
         need = n%2==0?n/2:n/2+1;
         int[] ar = new int[n];

         for(int i = 0;i < n;i ++)
             ar[i] = in.nextInt();
         System.out.println(Partition(ar, 0, n-1));

     }
 }

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。