二分查找法

#include <iostream>
using namespace std;

 

//法一

int bfind(int a[],int len,int val)
{
 int end = len/2;    //中间位置
 int start = 0;      //开始位置
 int rLen = len;     //剩余检测数组的长度
 while(start != end && rLen != end)
 {
  if(val < a[end])
  {
   rLen = end;
   end = (end+start)/2;
  }
  else if(val > a[end])
  {
   start = end;
   end = (end+rLen)/2;
  }
  else
   return end;
 }
 return -1;
}

 

// 法二（这个很好理解）

int binarySearh(int a[],int len,int x) 
{
 int first = 0;
 int last = len-1;
 int mid;
 while(first <= last)
 {
  mid = first + (last-first)/2; //直接使用(high+first)/2会有整数溢出的情况发生
  if(x == a[mid])
   return mid;
  else if(x < a[mid])
   last = mid-1;
  else
   first = mid+1;
 }
 return -1;
}

int main()
{
 int a[] = {1,3,4,5,7,8,9,11,17,38,56};
 int b;
 cin>>b;
 //int count = bfind(a,11,b);
 int count = binarySearh(a,11,b);
 if(count == -1)
  cout<<"no reslution!"<<endl;
 else
  cout<<count<<endl;
 return 0;
}