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;
}