模板 #

以 Unix 风格给出一个文件的绝对路径，你需要简化它。或者换句话说，将其转换为规范路径。 在 Unix 风格的文件系统中，一个点（.）表示当前目录本身；此外，两个点 （..）  表示将目录切换到上一级（指向父目录）；两者都可以是复杂相对路径的组成部分。更多信息请参阅：Linux / Unix 中的绝对路径 vs 相对路径 请注意，返回的规范路径必须始终以斜杠 / 开头，并且两个目录名之间必须只有一个斜杠 /。最后一个目录名（如果存在）不能以 / 结尾。此外，规范路径必须是表示绝对路径的最短字符串。 示例 1： ``` 输入："/home/" 输出："/home" 解释：注意，最后一个目录名后面没有斜杠。 ``` 示例 2： ``` 输入："/../" 输出："/" 解释：从根目录向上一级是不可行的，因为根是你可以到达的最高级。 ``` 示例 3： ``` 输入："/home//foo/" 输出："/home/foo" 解释：在规范路径中，多个连续斜杠需要用一个斜杠替换。 ``` 示例 4： ``` 输入："/a/./b/../../c/" 输出："/c" ``` 示例 5： ``` 输入："/a/../../b/../c//.//" 输出："/c" ``` 示例 6： ``` 输入："/a//b////c/d//././/.." 输出："/a/b/c" ```

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. 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" ```

先按 `/` 进行切割，然后创建一个 `stack`，判断为 `.` 就跳过，为 `..` 就 `pop`（`stack` 非空），其他就 `append`

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```python class Solution: def simplifyPath(self, path: str) -> str: ps = [p for p in path.split('/') if p] stack = list() for p in ps: if p == '.': continue elif p == '..': if stack: stack.pop() else: stack.append(p) return '/' + '/'.join(stack) ```

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