7递归二分和迭代二分 

  1 /*
  2     递归二分查找_I
  3     迭代二分查找_R
  4 
  5 */
  6 
  7 
  8 #include "BinarySearch_I_and_R.h"
  9 #include <iostream>
 10 using namespace std;
 11 int BinarySearch_I(int *a,int nLen,int nTarget);
 12 int BinarySearch_R(int *a,int nTarget,int left,int right);
 13 
 14 int main()
 15 {
 16     int nTest=7;
 17     const int nLen = 10;
 18     int m[nLen] = {1,2,3,4,5,6,7,8,9,10};
 19 
 20     int nRet = BinarySearch_I(m,nLen,nTest);
 21     if (-1 == nRet)
 22     {
 23         cout <<"迭代二分:没找到"<<endl;
 24     }
 25     else
 26     {
 27         cout <<"迭代二分:找到了"<<endl;
 28     }
 29 
 30 
 31     nRet = BinarySearch_R(m,nTest,0,nLen-1);
 32     if (-1 == nRet)
 33     {
 34         cout <<"递归二分:没找到"<<endl;
 35     }
 36     else
 37     {
 38         cout <<"递归二分:找到了"<<endl;
 39     }
 40 
 41 
 42     
 43 
 44     system("pause");
 45     return 0;
 46 }
 47 
 48 
 49 //迭代二分
 50 int BinarySearch_I(int *a,int nLen,int nTarget)
 51 {
 52     int left = 0;
 53     int right = nLen-1;
 54     int mid;
 55 
 56     while(left <= right)
 57     {
 58         mid = (left + right) / 2;
 59         if (nTarget == a[mid])
 60         {
 61             return mid;
 62         }
 63         else if (nTarget > a[mid])
 64         {
 65             left = mid + 1;
 66         }
 67         else if (nTarget < a[mid])
 68         {
 69             right = mid - 1;
 70         }
 71 
 72     }
 73 
 74     return -1;
 75 }
 76 
 77 
 78 //递归二分
 79 int BinarySearch_R(int *a,int nTarget,int left,int right)
 80 {
 81 
 82     if (left <= right)
 83     {
 84         int mid = (left + right)/2;
 85         if (nTarget < a[mid])
 86         {
 87             return BinarySearch_R(a,nTarget,left,mid-1);
 88         }
 89         else if (nTarget > a[mid])
 90         {
 91             return BinarySearch_R(a,nTarget,mid+1,right);
 92         }
 93         else
 94         {
 95 
 96             return mid;
 97         }
 98     }
 99 
100 
101     return -1;
102 }

 

posted @ 2017-05-05 15:16  w6811645  阅读(255)  评论(0)    收藏  举报