原题地址：https://oj.leetcode.com/problems/simplify-path/

题意：

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".

解题思路：

题目的要求是输出Unix下的最简路径，Unix文件的根目录为"/"，"."表示当前目录，".."表示上级目录。

例如：

输入1：

/../a/b/c/./.. 

输出1：

/a/b

模拟整个过程：

1. "/" 根目录

2. ".." 跳转上级目录，上级目录为空，所以依旧处于 "/"

3. "a" 进入子目录a，目前处于 "/a"

4. "b" 进入子目录b，目前处于 "/a/b"

5. "c" 进入子目录c，目前处于 "/a/b/c"

6. "." 当前目录，不操作，仍处于 "/a/b/c"

7. ".." 返回上级目录，最终为 "/a/b"

使用一个栈来解决问题。遇到'..'弹栈，遇到'.'不操作，其他情况下压栈。

代码一：

class Solution:    # @param path, a string    # @return a string    def simplifyPath(self, path):        stack = []        i = 0        res = ''        while i < len(path):            end = i+1            while end < len(path) and path[end] != "/":                end += 1            sub=path[i+1:end]            if len(sub) > 0:                if sub == "..":                    if stack != []: stack.pop()                elif sub != ".":                    stack.append(sub)            i = end        if stack == []: return "/"        for i in stack:            res += "/"+i        return res

代码二：

利用python的字符串处理能力。

class Solution:    # @param path, a string    # @return a string    def simplifyPath(self, path):        path = path.split('/')        curr = '/'        for i in path:            if i == '..':                if curr != '/':                    curr = '/'.join(curr.split('/')[:-1])                    if curr == '': curr = '/'            elif i != '.' and i != '':                curr += '/' + i if curr != '/' else i        return curr