寻找数组N中最大（最小的）M个数（亲自调试可运行）

标签：blog   io   使用   ar   for   数据   2014   art   sp   

当N很小十可以使用方法2，

当N很大时可以使用方法1，从硬盘逐次读入解决；

 *

*/

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define NDEBUG
#include <assert.h>

int  input(void);
void print(int * p,int len);
void adjust(int *heap,int heap_size,int index);
void swap(int *a,int *b);
void built_heap(int *heap,int heap_size,int *a);
void find_max(int *heap,int heap_size,int *a,int len);
//方法2
void  quick_find(int *arry,int len, int M);
int partation(int *arry,int len,int start,int end);

int main(void)
{
  int heap_size;
  int *heap;
  //int a[]={23,54,47,12,6,23,14,1,25,3,5,4,9};
  int a[]={1,231,45,69,78,15,36,49,14,36,24,15,89,12,11,69,122,192,126,2125,1,12,4548,125,4755,232,            5,26,24,25,22,22,66,214,1235,2,332,56,66,662,552,225,226,226,2262,262,626,2626,2626,789,         622,15,25,262,551,789,781,895,895,9511,942,412,233,203,202,30,95,2,31,22,12,23,325,235,266,45678,15629,7894};
  heap_size=input();
  heap=(int *)malloc(sizeof(int)*heap_size);
  find_max(heap,heap_size,a,sizeof(a)/sizeof(int));
  print(heap,heap_size);
  free(heap);
  quick_find(a,sizeof(a)/sizeof(int), heap_size);
  return 0;
} 

void print(int * p,int len)
{
    int i;
    for(i=0;i<len;i++)
      printf("%d  ",p[i]);
      printf("\n"); 
    return;  
}

int  input(void)
{
    int m;
    printf("input the M:");
    scanf("%d",&m);
    return m;
}

//建立堆
void built_heap(int *heap,int heap_size,int *a)
{
    int i;
    assert(heap!=NULL&&a!=NULL);
    for(i=0;i<heap_size;i++)
         heap[i]=a[i];
    for(i=(heap_size>>1);i>=0;i--)
       adjust(heap,heap_size,i);
    return;         
}
//堆调整
void adjust(int *heap,int heap_size,int index)
{
   int left=(index<<1)+1;
   int right=(index<<1)+2;
   int parent=index;
   assert(heap!=NULL);
   if(left<heap_size&&heap[left]<heap[parent])
            parent=left;     
   if(right<heap_size&&heap[right]<heap[parent])
            parent=right;
   if(parent!=index)
     {
        swap(&heap[parent],&heap[index]);
        adjust(heap,heap_size,parent);
     }             
    return;  
} 


void swap(int *a,int *b)
{
    *a=*a^*b;
    *b=*a^*b;
    *a=*a^*b;
}

void find_max(int *heap,int heap_size,int *a,int len)
{
     int i;
     int j;
     assert(heap!=NULL&a!=NULL);
     if(heap_size>=len)
     return;
     built_heap(heap,heap_size,a);
     for(i=heap_size;i<len;i++)
     {
         if(a[i]>heap[0])
            {
               heap[0]=a[i];
               adjust(heap,heap_size,0);
            }   
     }
     return;
}
//方法2
void  quick_find(int *arry,int len, int M)
{
     int start=0;
     int end=len-1;
     int count=M-1;
     int i;
     int proe;
     if(arry==NULL||len<M)
     return;
     proe=partation(arry,len-1,start,end);
     while(proe!=count)
     {
         if(proe>count)
            proe=partation(arry,len-1,start,proe-1);
         else
            proe=partation(arry,len-1,proe+1,end);
     }
     for(i=0;i<M;i++)
     printf("%d  ",arry[i]);
     printf("\n");
     
}
int partation(int *arry,int len,int start,int end)
{
     int key=arry[start];
     while(start<end)
     {
          while(start<end&&arry[end]<=key)
               end--;
           if(start<end)
              arry[start++]=arry[end];    
          while(start<end&&arry[start]>key)
              start++;
          if(start<end)
             arry[end--]=arry[start];         
     }
     arry[start]=key;
     return start;
}

程序运行结果：

“输出两行分别为方法1和方法2结果”

[trageday@lei-yum code_test]$ gcc   -o arryN_maxM arryN_maxM.c
[trageday@lei-yum code_test]$ ./arryN_maxM 
input the M:10
2125  2262  2626  4755  2626  45678  9511  7894  15629  4548  
45678  15629  9511  7894  4548  2626  2626  4755  2262  2125 

寻找数组N中最大（最小的）M个数（亲自调试可运行）

标签：blog   io   使用   ar   for   数据   2014   art   sp   

原文地址：http://blog.csdn.net/trageday_motata/article/details/39099103