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