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）