71. Simplify Path(Python3)

xiaoxiao2021-02-28  2

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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