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：

class Solution:
    # @param path, a string
    # @return a string
    def simplifyPath(self, path):
        pathSplit=path.split("/")
        stack=[]
        res=""
        for i in range(len(pathSplit)):
            if pathSplit[i] =="." or len(pathSplit[i])==0 :
                continue;
            elif pathSplit[i] =="..":
                if len(stack)!=0:
                    stack.pop()
            else: stack.append(pathSplit[i])
        if len(stack)==0:
            stack.append("")
        while len(stack)!=0:
            res+="/"+stack.pop(0)
        return res
        

简化版：

class Solution:
    # @param path, a string
    # @return a string
    def simplifyPath(self, path):
        pathSplit=path.split("/")
        stack=[]
        for ps in pathSplit:
            if ps !="." and ps!=".." and ps :
                stack.append(ps)
            elif ps ==".." and stack :
                    stack.pop()
        return "/"+"/".join(stack)