二分法(折半查找)

(1)直接二分法查找
#include<stdio.h>
int main()
{
 int arr[]={1,2,3,4,5,6,7,8,9};
 int k = 7;
 int sz = sizeof(arr)/sizeof(arr[0]);
 int left = 0;
 int right = sz-1;
 while(left<=right)
 {
  int mid = left + (right-left)/2;
        if(k==mid)
  {
            printf("找到了:%d\n",mid); 
   break;
  }
  else if(k<mid)
  {
          right=mid-1;  
  }
  else
  {
         left=mid+1;  
  }
 }
 if(left>right)
 {
 printf("找不到\n");
 }
return 0;
}
(2)对函数进行调用
#include<stdio.h>
void binary_search(int arr[],int k,int sz)
{
 int left = 0;
 int right = sz-1;
   while(left<=right)
 {
  int mid = left + (right-left)/2;
        if(k==mid)
  {
            printf("找到了:%d\n",mid); 
   break;
  }
  else if(k<mid)
  {
          right=mid-1;  
  }
  else
  {
         left=mid+1;  
  }
 }
 if(left>right)
 {
 printf("找不到\n");
 }
}
int main()
{
 int arr[]={0,1,2,3,4,5,6,7,8,9,};
 int k = 7;
 int sz = sizeof(arr)/sizeof(arr[0]);
 binary_saerch(&arr[0],k,sz);
return 0;
}
(3) 对函数进行优化
#include<stdio.h>
int  binary_search(int arr[],int k,int sz)
{
int left =0;
  int right=sz-1;
   while(left<=right)
   {
    int mid=left+(right-left)/2;
       if(arr[mid]==k)
       { 
          return mid;
       }
       else if(arr[mid]>k)
       {
           right=mid-1;
       }
      else
      {
          left=mid+1;
      }
   }
   return -1;
}
int main()
{
 int arr[]={1,2,3,4,5,6,7,8,9};
 int k =7;
 int sz=sizeof(arr)/sizeof(arr[9])-1;
 int ret = binary_search(arr,k,sz);
 if(ret==-1)
 {
    printf("找不到\n");
 }
 else
 {
    printf("找到了 %d\n",ret);
 } 
 return 0;
}


猜你喜欢

转载自blog.csdn.net/dreamispossible/article/details/79773860