Given an absolute path for a file (Unix-style), simplify it.
For example,
path = "/home/"
, => "/home"
path = "/a/./b/../../c/"
, => "/c"
path = "/a/../../b/../c//.//"
, => "/c"
path = "/a//b////c/d//././/.."
, => "/a/b/c"
In a UNIX-style file system, a period ('.') refers to the current directory, so it can be ignored in a simplified path. Additionally, a double period ("..") moves up a directory, so it cancels out whatever the last directory was. For more information, look here: https://en.wikipedia.org/wiki/Path_(computing)#Unix_style
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"
.
结合条件来看,/
除了用作分隔符之外没有用处;.
表示当前dir,也没有用;..
表示返回上一个dir,从stack里面pop出最后一个,如果空的话就忽略(我觉得实际中不应该忽略的emm)。其余情况就push入栈。
最后用/
把栈里面的东西连起来即可。
class Solution:
def simplifyPath(self, path):
"""
:type path: str
:rtype: str
"""
list = path.split('/')
stack = []
output = ''
for c in list:
if len(c) == 0 or c == '.':
continue
elif c == '..':
if len(stack) is not 0:
stack.pop()
else:
stack.append(c)
if len(stack) == 0:
output = '/'
else:
for c in stack:
output += '/' + c
# print(output)
return output