71 Simplify Path

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".
The Idea: This problem requires some Unix domain knowledge. . means current directory, so in terms of a path, it adds nothing and can be ignored. .. means move up to parent directory, and ... means move up recursively. Utilizing a stack is a natural solution because moving to a parent directory is equivalently to moving to the previous or most recent element in an array.
Complexity: O(n) time and space
class Solution:
    def simplifyPath(self, path):
        """
        :type path: str
        :rtype: str
        """
​
        tokens = [token for token in path.split('/') if token != '' and token != '.']
        s = []
​
        for token in tokens:
            if token == '..':
                if s: # separated ifs for case where '...'
                    s.pop()
            else:
                s.append(token)
​
        return '/' + '/'.join(s)
​
​
obj = Solution()
print(obj.simplifyPath('/...')) # /...
print(obj.simplifyPath('/../')) # /
print(obj.simplifyPath('/')) # /
print(obj.simplifyPath('/a/./b/../../c/')) # /c
print(obj.simplifyPath('/home//foo/')) # /home/foo
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Last modified 4yr ago