[Stack]71. Simplify Path

• 分类：Stack
• 时间复杂度: O(n)
• 空间复杂度: O(n)

71. Simplify Path

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. For more information, see: Absolute path vs relative path in Linux/Unix

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"

代码：

class Solution:
    def simplifyPath(self, path: 'str') -> 'str':
        
        stack=[]
        
        paths=path.split("/")
        for p in paths:
            if p=="..":
                if stack!=[]:
                    stack.pop()
            elif p!="." and p!="":
                stack.append(p)
        
        res="/"
        if stack==[]:
            return res
        for s in stack:
            res+=s+"/"
        return res[:-1]

讨论：

1.这道题挺简单的，用stack方法就能完美解决
2.凡是遇到这种“路径类”的问题都用stack解，算是一个固定套路