[LeetCode] Simplify Path

最新推荐文章于 2019-05-21 14:44:54 发布

a83610312

最新推荐文章于 2019-05-21 14:44:54 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/a83610312/article/details/10033363

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本文介绍了一个简化Unix文件系统中绝对路径的算法实现。通过使用栈结构处理路径中的符号（如'.'、'..'），该算法能有效解决路径规范化问题，并处理特殊情况，如连续斜杠及上级目录引用。

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Simplify Path:

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

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

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".

如果想按字符遍历的话就比较烦了，所以不如按'/' 斜杠来分割，然后按词来处理，就变成一个简单的栈的处理了。

class Solution {
public:
    string simplifyPath(string path) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function

		vector<string> st;
		string label;
		int start=0;
		while(getLabel(path,start,label))
		{
			if ( label.compare("/") ==0 )
			{
				if ( st.empty() || st.back().compare("/")!=0)
					st.push_back(label);
			}
			else if ( label.compare("..")==0)
			{
				assert(!st.empty()&&st.back().compare("/")==0);
				st.pop_back();
				if (!st.empty())
					st.pop_back();
			}
			else if ( label.compare(".")==0)
			{
				assert(!st.empty()&&st.back().compare("/")==0);
				st.pop_back();
			}
			else
			{
				assert(!st.empty()&&st.back().compare("/")==0);
				st.push_back(label);
			}
		}
		string ret;
        vector<string>::iterator it;
    	for(it=st.begin();it!=st.end();++it)
			ret.append(*it);
            
        if (ret.length()>0 && ret.back()=='/')
            ret.pop_back();
                
		if ( ret.empty() )
			ret.append("/");

		return ret;
	}
	bool getLabel(string& path,int& start, string& label)
	{
		int len=path.length();
		if ( start==len)
			return false;
		if ( path[start]=='/')
		{
			label="/";
			start++;
			return true;
		}
		int end =start+1;
		while( end<len)
		{
			if(path[end]=='/')
				break;
			end++;
		}
		label=path.substr(start,end-start);
		start=end;
		return true;
	}
};

之前还自己写个分割的函数，太Naive了，直接用库函数find就好了呀：

class Solution {
public:
    string simplifyPath(string path) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector<string> st;
        string::iterator it=path.begin();
        string::iterator pre=it;
        while((it=find(it,path.end(),'/'))!=path.end())
        {
            string tmp(pre,it);
            if(tmp.empty()||tmp==".")
            {
            }
            else if (tmp=="..")
            {
                if(!st.empty())
                    st.pop_back();
            }
            else
                st.push_back(tmp);
            pre=++it;
        }
                
        string ans;
        for(int i=0;i<st.size();i++)
        {
            ans.push_back('/');
            ans.append(st[i]);
        }
        if(ans.empty())
            return "/";
        else
            return ans;
    }
};