Leetcode's simplify Path @ python

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

For example,
Path = "/home/" , = ="/home"
Path = "/a/./b/../../c/" , = ="/c"

Use a stack to solve the problem. Encounter ' ... ' Stack, encounter '. ' Do not operate, other cases of pressure stack.

Code One:

classSolution: # @param path, astring# @return Astringdef simplifypath (self, path): Stack=[] I=0Res="'         whilei<len (path): End= i+1             whileEnd<len (path) and path[end]! ="/": End+=1Sub= path[i+1: end]ifLen (sub) >0:                ifSub = ="..":                    ifStack! =[]: Stack.pop () elif Sub!=".": Stack.append (sub) I=Endifstack = = []:             return "/"         forIinchStack:res+="/"+IreturnRes

Code 2:

classSolution:defSimplifypath (Self,path): Path= Path.split ('/') Res='/'         forIinchPath:ifi = ='..':                ifRes! ='/': Res='/'. Join (Res.split ('/') [:-1])                    ifres = ="': res ='/'            elifI! ='.'  andI! ="': Res+='/'+iifRes! ='/' ElseIreturnRes

Turn from (reference):

1. http://www.cnblogs.com/zuoyuan/p/3777289.html

2. http://blog.csdn.net/linhuanmars/article/details/23972563

@ JAVA Version

Public string Simplifypath (string path) {if(Path = = NULL | | path.length () = =0) {return ""; } LinkedList<String> stack = new linkedlist<string>(); StringBuilder Res=new StringBuilder (); int i=R;  while(i<path.length ()) {int index=i; StringBuilder Temp=new StringBuilder ();  while(I<path.length () && Path.charat (i)! ='/') {temp.append (Path.charat (i)); I++; }        if(index!=i) {String str=temp.tostring (); if(Str.equals (".."))            {                if(!stack.isempty ()) Stack.pop (); }            Else if(!str.equals (".") ) {Stack.push (str); }} I++; }    if(!stack.isempty ()) {string[] STRs=Stack.toarray (New String[stack.size ()]);  for(int j=strs.length-1;j>=0;j--) {res.append ("/"+Strs[j]); }    }    if(res.length () = =0)return "/"; returnres.tostring ();}

Leetcode's simplify Path @ python