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：
![[leetcode]Simplify Path @ Python](/default/index/img?u=aHR0cDovL2NvbW1vbi5jbmJsb2dzLmNvbS9pbWFnZXMvY29weWNvZGUuZ2lm "[leetcode]Simplify Path @ Python")
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