binary-search( 二分查找 )

#ifndef _BINARYSEARCH_H_#define _BINARYSEARCH_H_template< class T >int binarysearch( T * lpArr, const T & v, int num ){//lpArr指向从小到大排列的数组int iStart = 0;int iEnd = num - 1;int iMid = 0;if( ( v > lpArr[ iEnd ] ) || ( v < lpArr[ iStart ] ) ){return -1;}while( true ){iMid = ( iStart + iEnd ) / 2;if( v == lpArr[ iMid ] ){return iMid;}if( v < lpArr[ iMid ] ){iEnd = iMid;}else {iStart = iMid + 1;}if( iStart == iEnd ){if( v == lpArr[ iStart ] ){return iStart;}else{return -1;}break;}}return -1;}#endif

这个例子是使用二分查找法在一个n个元素的集合里面查找是否存在两个和为10的元素，是的话在控制台打印“find！“这是《算法导论》中的题目，符合题目要求，使最坏情况下运行时间为O（ nlgn ）

#include<iostream>using namespace std;#include<conio.h>#include"binarysearch.h"int main(){int a[ 8 ] ={ 2, 4, 5, 6, 4, 56, 77, 89 };for( int i = 0; i < 8; ++i ){cout<<a[ i ]<<" ";}cout<<endl;for( int i = 0; i < 8; ++i ){int temp = 10 - a[ i ]; if( binarysearch( a, temp, 8 )>= 0 ){cout<<"find!"<<endl;break;}}getch();return 0;}