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