二分查找那些事儿(1)
2013-06-08 09:47
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#include <iostream> using namespace std; /*以下都为正确的二分查找算法*/ //左闭右闭[left,right] int bisearch1(int arr[],int n,int v) { int left,right,mid; left=0; right=n-1; while(left<=right) //right=left-1时退出循环,且一定找不到v { mid=left+(right-left)/2; if(arr[mid]<v) left=mid+1; else if(arr[mid]>v) right=mid-1; else return mid; } return -1; } //左闭右开[left,right) int bisearch2(int arr[],int n,int v) { int left,right,mid; left=0; right=n; while(left<right) //left==right时退出循环,且arr[right]一定不等于v,所以也一定找不到v { mid=left+(right-left)/2; if(arr[mid]<v) left=mid+1; else if(arr[mid]>v) right=mid; else return mid; } return -1; } //左开右开(left,right) int bisearch3(int arr[],int n,int v) { int left,right,mid; left=-1; right=n; while(left+1!=right) //left+1=right时退出循环,且arr[left],arr[right]一定不等于v,所以也一定找不到v { mid=left+(right-left)/2; if(arr[mid]<v) left=mid; else if(arr[mid]>v) right=mid; else return mid; } return -1; } /* 存在不足:以上三个算法都只能返回任意一个i,使得arr[i]==v,不能返回第一个arr[i]==v的位置 循环体中分三种情况,两次比较,可以减小比较次数优化程序; 改进请看下篇 */ int main() { int array[]={1,1,2,3,4,4,5,6,7,8,8,9,10}; int len=sizeof(array)/sizeof(int); int value; value=4; cout<<bisearch1(array,len,value)<<endl; cout<<bisearch2(array,len,value)<<endl; cout<<bisearch3(array,len,value)<<endl; return 0; }