LeetCode 071 Simplify Path

LeetCode

Given a string path, which is an absolute path (starting with a slash '/') to a file or directory in a Unix-style file system, convert it to the simplified canonical path.

In a Unix-style file system, a period '.' refers to the current directory, a double period '..' refers to the directory up a level, and any multiple consecutive slashes (i.e. '//') are treated as a single slash '/'. For this problem, any other format of periods such as '...' are treated as file/directory names.

The canonical path should have the following format:

The path starts with a single slash '/'. Any two directories are separated by a single slash '/'.
The path does not end with a trailing '/'. The path only contains the directories on the path from the root directory to the target file or directory (i.e., no period '.' or double period '..')

Return the simplified canonical path.

Example 1:

Input: path = “/home/“
Output: “/home”
Explanation: Note that there is no trailing slash after the last directory name.

Example 2:

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

Constraints:

1 <= path.length <= 3000 path consists of English letters, digits, period '.', slash '/' or '_'.
* path is a valid absolute Unix path.

题目大意：

简化路径，遇.表示当前目录不做事，遇..表示到上一个目录

解题思路：

路径类似于括号题，利用括号题模板

解题步骤：

N/A

注意事项：

1. edge case /../ 表示若stack为空，就不pop。if stack不能加到elif token == ‘..’中
2. 遇到..返回到上层目录

Python代码：

def simplifyPath(self, path: str) -> str:
	path += '/'
	token, stack = '', []
	for c in path:
		if c == '/':
			if token == '.':
				pass
			elif token == '..':
				if stack: # remember
					stack.pop()
			elif token:
				stack.append(token)
			token = ''
		else:
			token += c
	return '/' + '/'.join(stack)

算法分析：

时间复杂度为O(n)，空间复杂度O(n)