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"

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".
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

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