中位数与顺序统计

 

#include <iostream>
#include <iomanip>
using namespace std;

 
void swap(int &x, int &y)
{
    int tmp = x;
    x = y;
    y = tmp;
}
 
 void insert_sort(int A[], int p, int r)
 {
     int i,j;
     int key = 0;
     for (i=p+1; i<=r; ++i)
     {         
         if(A[i]<A[i-1])
         {
             key = A[i];
             for(j=i-1;j>=p && A[j]>key;--j)
             {
                 A[j+1] = A[j];
             }
             A[j+1] = key;
         }        
     }
 }
 
 
 int partion(int A[],int p, int r)
 {
     int count = p - 1;
     int key = A[r]; 
     for (int i=p; i<=r-1; i++)
     {
         if (A[i] < key)
         {
             count++;
             swap(A[count],A[i]);
         }
     }
     swap(A[count+1],A[r]);
     return count+1;
 }
 
 int get_median(int data[], int p, int r)
 {
     int i=0, j=0;
     int n = r - p + 1;           
     int remains = n%5;
     int int_count = n - remains;
     int int_group = int_count/5; 
     int group_count = (n+4)/5;   
     
     int *group_median = new int[group_count];
     
     if (p==r)
     {
         return data[p];
     }
     memset(group_median,0,group_count);      //对动态申请的内存清零
     
     //以下代码求出五个元素为一组的中位数
     if (0 == remains) //如果元素的个数正好可以分为以5个元素为单位的整数组
     {
         for (i=p; i<=r-4; i=i+5)
         {
             insert_sort(data, i, i+4);
             group_median[(i-p)/5] = data[i+2];
         }
     }
     else
     {
         for (i=p; i<=(r-remains-4);i=i+5)
         {
             insert_sort(data, i, i+4);
             group_median[(i-p)/5] = data[i+2];
         }
         //处理不够5个元素的组，改组开始的序号为r-remains+1，结束序号为r
         insert_sort(data,r-remains+1,r);
         group_median[group_count-1] = data[r-remains+1+remains/2];
     }
 

     if (group_count==1)
     {
         return group_median[0];
     }
     else
     {
         return get_median(group_median,0,group_count-1);
     }
 
     delete [] group_median;
     return 0;
 }
 

 int get_position(int A[], int p, int r, int key)
 {
     for (int i=p; i<=r; i++)
     {
         if (A[i]==key)
         {
             return i;
         }
     } 
     return -1;
 }
 
 //该函数是为了找到数组A中，第seq小的数
 int select(int A[],int p, int r, int seq)
 {
     //获得数组A的中位数的中位数，将其作为划分数组的支点
     int pivot_key = get_median(A, p, r);
     int pos = get_position(A,p,r,pivot_key);
     swap(A[pos],A[r]);

     int i = partion(A, p, r); 





     int x = i - p + 1;
     if (seq == x)
     {
         return A[i];
     }
     else if (seq < x)
     {
         select(A, p, i-1, seq);
     }
     else
     {
         select(A, i+1, r, seq-x);
     }
 
 }
 
 
 int main()
 {
     int arrays[] = {23,13,12,14,11,8,9,20,7,4,5,6,2,1,3};
     int size = sizeof(arrays) / sizeof(int);
 
     for (int i=1; i<=size; i++)
     {
             cout << "select "<<i<<"->"<< select(arrays,0,size-1,i) << endl; 
     }

     char wait;
     cin>>wait;

 
     return 0;
 }