首页 > 其他 > 详细

二分查找

时间:2016-02-25 15:40:22      阅读:296      评论:0      收藏:0      [点我收藏+]

FROM:

http://blog.csdn.net/int64ago/article/details/7425727

对于YES_LEFT或者NO_RIGHT

int bSearch(int begin, int end, int e)  
{  
    int mid, left = begin, right = end;  
    while(left <= right)  
    {  
        mid = (left + right) >> 1;  
        if(num[mid] >= e) right = mid - 1;  
        else left = mid + 1;  
    }  
    return left;  
}  

对于YES_RIGHT或者NO_LEFT

int bSearch(int begin, int end, int e)  
{  
    int mid, left = begin, right = end;  
    while(left <= right)  
    {  
        mid = (left + right) >> 1;  
        if(num[mid] > e) right = mid - 1;  
        else left = mid + 1;  
    }  
    return right;  
}  

 

二分查找

原文:http://www.cnblogs.com/tun117/p/5217005.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!