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71. Simplify Path(Python3)

曹智
2023-12-01

71. Simplify Path(Python3)

题目

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”.

解题方案

思路:

  • 本题主要是对目录的处理,其实并不复杂,学习到了如何去掉list中的空字符,之前一直在用filter来做,但是filter其实写起来还要在外层加一个list,因为python3直接返回值不是list
  • 除此之外,就是有的解法将list当作stack来使用感觉很巧妙

代码:

class Solution:
    def simplifyPath(self, path):
        """
        :type path: str
        :rtype: str
        """
        lpath = [v for v in path.split("/") if v != "." and v != ""]
        res = []
        for i in range(len(lpath)):
            if lpath[i] == '..':
                if i-1 >= 0:
                    res = res[:-1]
            else:
                res.append(lpath[i])
        return '/'+'/'.join(res)
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