Simplify Path

最新推荐文章于 2020-02-07 13:43:23 发布

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最新推荐文章于 2020-02-07 13:43:23 发布

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分类专栏： c++/c

本文链接：https://blog.youkuaiyun.com/njust_ecjtu/article/details/37613103

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本文介绍了一种简化Unix文件路径的方法，通过解析输入的绝对路径并处理各种特殊情况，如多个连续的斜杠、相对路径符号等，最终返回规范化的路径。

Given an absolute path for a file (Unix-style), simplify it.

For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"

click to show corner cases.

Corner Cases:

Did you consider the case where path = "/../"?
In this case, you should return "/".
Another corner case is the path might contain multiple slashes '/' together, such as "/home//foo/".
In this case, you should ignore redundant slashes and return "/home/foo".

此题的很多特殊情况，此处代码写的不是很好！

为 . .. 字母目录，（特别是为空的情况，在此次使用了一个back_root 标志位，并且下一次还是为/重复，去重！）

返回只有在/..才返回！../ 输出为..

string simplifyPath(string path) 
{
	if(path.empty())return path;

	int pos = 0;
	string res;
	bool back_root = false;
	while(pos < path.size())
	{
		if(path[pos] == '/' || back_root)
		{
			back_root = false;
			if(path[pos+1] == '\0' )//处理要结束的情况
			{
				break;
			}
			else if(pos +3 <= path.size()&&path[pos+1] == '.' &&path[pos+2] == '.'&&(path[pos+3] == '/'||path[pos+3] == '\0'))//..的情况

			{
				if(!res.empty())
				{
					res = res.substr(0,res.rfind('/'));//上一个地方
				}
				else
				{
					back_root = true;
					res = "/";
				}
				pos = pos + 3;
			}
			else if(pos +2 <= path.size()&&path[pos+1] == '.'&&(path[pos+2] == '/'||path[pos+2] == '\0'))// .的情况 

			{
				pos = pos + 2;
			}
			else if((res.size() >= 1&&res[res.size()-1]=='/')||path[pos+1] == '/') //处理 /// 多个情况，以及返回为空res多一个/也要在此处处理
			{
				pos++;
			}
			else//处理字母目录
			{
				
				int start = pos;
				pos++;
				while(path[pos] != '/'&&path[pos] !='\0')
				{
					pos++;
				}
				res = res + path.substr(start,pos-start);
			}
		}
		else  //处理 非/的情况！！全部加入
		{
			res += path[pos];
			pos++;
		}


	}
	if(res.empty())res = "/";
	return res;
}

第二次：

用一个vcetor模拟栈的功能，只存储字符串。/后续的补上

string simplifyPath(string path) 
{
	if(path.empty())return path;

	vector<string> path_vec;
	
	int pos = 0;
	while(pos < path.size()&& pos != -1 )
	{
		if(path[pos] == '/')
		{
			pos++;
			 
		    int pos_end = path.find('/',pos);
			string sub_str ;

			if(pos_end == -1)//未找到
			{
				sub_str = path.substr(pos);
			}
			else
			{
				sub_str = path.substr(pos,pos_end-pos);
			}
			

			if (sub_str == ".." )
			{
				if(!path_vec.empty())
				{
					path_vec.pop_back();
				}
				pos = pos_end;
			}
			else if (sub_str == "."||sub_str == "")//只需后面移动位置,sub_str == ""为/
			{
			  pos = pos_end;
			}
			else
			{
				path_vec.push_back(sub_str);
				pos = pos_end;
			}
			
		}
		
	}
	string res;

	for (int i = 0;i < path_vec.size();i++)
	{
		res +="/"+ path_vec[i];
	}
	if (path_vec.empty())
	{
		res = "/";
	}
	return res;
}