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71. Simplify Path-LeetCode-Pytho

71. Simplify Path-LeetCode-Pytho

作者: 云外雁行斜丶 | 来源:发表于2019-06-08 08:50 被阅读0次

    71. Simplify Path

    Given an absolute path for a file (Unix-style), simplify it. Or in other words, convert it to the canonical path.

    In a UNIX-style file system, a period . refers to the current directory. Furthermore, a double period .. moves the directory up a level. For more information, see: Absolute path vs relative path in Linux/Unix

    Note that the returned canonical path must always begin with a slash /, and there must be only a single slash / between two directory names. The last directory name (if it exists) must not end with a trailing /. Also, the canonical path must be the shortest string representing the absolute path.

    Example 1:

    
    Input: "/home/"
    Output: "/home"
    Explanation: Note that there is no trailing slash after the last directory name.
    

    Example 2:

    
    Input: "/../"
    Output: "/"
    Explanation: Going one level up from the root directory is a no-op, as the root level is the highest level you can go.
    

    Example 3:

    
    Input: "/home//foo/"
    Output: "/home/foo"
    Explanation: In the canonical path, multiple consecutive slashes are replaced by a single one.
    

    Example 4:

    
    Input: "/a/./b/../../c/"
    Output: "/c"
    

    Example 5:

    
    Input: "/a/../../b/../c//.//"
    Output: "/c"
    

    Example 6:

    
    Input: "/a//b////c/d//././/.."
    Output: "/a/b/c"
    

    Firstly

    先将路径切割成一个个单词
    根据栈的规律,判断是否是..或者.进行筛选,然后拼接

    class Solution(object):
        def simplifyPath(self, path):
            """
            :type path: str
            :rtype: str
            """
            path_lst = [s for s in path.split('/') if s.strip()]
            stack = []
            for s in path_lst:
                if s == '.':
                    continue
                if s == '..':
                    if stack:
                        stack.pop()
                else:
                    stack.append(s)
            return '/' + '/'.join(stack)
    
    

    Fastest

    思想一样

    class Solution(object):
        def simplifyPath(self, path):
            """
            :type path: str
            :rtype: str
            """
            stack = []
            for item in path.split('/'):
                if not item: continue
                if item=='..': 
                    if stack: stack.pop()
                elif item != '.': stack.append(item)
            return '/'+'/'.join(stack)
    

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