leetcode------Simplify Path

标题：	Simplify Path
通过率：	20.1%
难度：	中等

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".

这个题要了解linux文件命令，“..”意思是返回上级，"."或者空""是本级的意思，

那么将path分片遇到"."或者""跳过，遇到".."弹栈，其他情况进栈，

还要一种情况就是path本身就是空，那么在整个循环结束够进栈一个空值。

代码如下：

 1 class Solution:

 2     # @param path, a string

 3     # @return a string

 4     def simplifyPath(self, path):

 5         pathSplit=path.split("/")

 6         stack=[]

 7         res=""

 8         for i in range(len(pathSplit)):

 9             if pathSplit[i] =="." or len(pathSplit[i])==0 :

10                 continue;

11             elif pathSplit[i] =="..":

12                 if len(stack)!=0:

13                     stack.pop()

14             else: stack.append(pathSplit[i])

15         if len(stack)==0:

16             stack.append("")

17         while len(stack)!=0:

18             res+="/"+stack.pop(0)

19         return res

20         

 

简化版：

 1 class Solution:

 2     # @param path, a string

 3     # @return a string

 4     def simplifyPath(self, path):

 5         pathSplit=path.split("/")

 6         stack=[]

 7         for ps in pathSplit:

 8             if ps !="." and ps!=".." and ps :

 9                 stack.append(ps)

10             elif ps ==".." and stack :

11                     stack.pop()

12         return "/"+"/".join(stack)