用栈的思想来解决这个问题，遇到当前路径就跳过，遇到上一个路径，就删除上一个路径，剩下的情况直接入栈即可。
class Solution:
def simplifyPath(self, path: str) -> str:
stack = list()
dirs = path.split("/")
for dir in dirs:
if not dir or dir == ".":
continue
if dir == "..":
if stack:
stack.pop()
else:
stack.append(dir)
return '/' + '/'.join(stack)
image.png
71. Simplify Path[https://leetcode-cn.com/problems/simpli...
71. 简化路径 https://leetcode-cn.com/problems/simplify-path/[...
用栈的思想来解决这个问题，遇到当前路径就跳过，遇到上一个路径，就删除上一个路径，剩下的情况直接入栈即可。
71. Simplify Path Given an absolute path for a file (Unix...
Given an absolute path for a file (Unix-style), simplify ...
leetcode[https://leetcode-cn.com/problems/simplify-path/]
71、Simplify PathGiven an absolute path for a file (Unix-s...
题目： 思路 解法
题目链接https://oj.leetcode.com/problems/simplify-path/ Descr...
https://leetcode.com/problems/simplify-path/根据linux的路径规则，...
本文标题：leetcode 71 Simplify Path
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