[leetcode]Simplify Path @ Python

原題地址：https://oj.leetcode.com/problems/simplify-path/

題意：

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".

解題思路：

題目的要求是輸出Unix下的最簡路徑，Unix文件的根目錄為"/"，"."表示當前目錄，".."表示上級目錄。

例如：

輸入1：

/../a/b/c/./.. 

輸出1：

/a/b

模擬整個過程：

1. "/" 根目錄

2. ".." 跳轉上級目錄，上級目錄為空，所以依舊處於 "/"

3. "a" 進入子目錄a，目前處於 "/a"

4. "b" 進入子目錄b，目前處於 "/a/b"

5. "c" 進入子目錄c，目前處於 "/a/b/c"

6. "." 當前目錄，不操作，仍處於 "/a/b/c"

7. ".." 返回上級目錄，最終為 "/a/b"

使用一個棧來解決問題。遇到'..'彈棧，遇到'.'不操作，其他情況下壓棧。

代碼一：

class Solution:
    # @param path, a string
    # @return a string
    def simplifyPath(self, path):
        stack = []
        i = 0
        res = ''
        while i < len(path):
            end = i+1
            while end < len(path) and path[end] != "/":
                end += 1
            sub=path[i+1:end]
            if len(sub) > 0:
                if sub == "..":
                    if stack != []: stack.pop()
                elif sub != ".":
                    stack.append(sub)
            i = end
        if stack == []: return "/"
        for i in stack:
            res += "/"+i
        return res

代碼二：

利用python的字符串處理能力。

class Solution:
    # @param path, a string
    # @return a string
    def simplifyPath(self, path):
        path = path.split('/')
        curr = '/'
        for i in path:
            if i == '..':
                if curr != '/':
                    curr = '/'.join(curr.split('/')[:-1])
                    if curr == '': curr = '/'
            elif i != '.' and i != '':
                curr += '/' + i if curr != '/' else i
        return curr