首页 > 代码库 > [leetcode]Simplify Path @ Python

[leetcode]Simplify Path @ Python

2024-07-07 15:11:21   227人阅读

原题地址：https://oj.leetcode.com/problems/simplify-path/

题意：

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".
解题思路：

题目的要求是输出Unix下的最简路径，Unix文件的根目录为"/"，"."表示当前目录，".."表示上级目录。

例如：

输入1：

/../a/b/c/./.. 

输出1：

/a/b

模拟整个过程：

1. "/" 根目录

2. ".." 跳转上级目录，上级目录为空，所以依旧处于 "/"

3. "a" 进入子目录a，目前处于 "/a"

4. "b" 进入子目录b，目前处于 "/a/b"

5. "c" 进入子目录c，目前处于 "/a/b/c"

6. "." 当前目录，不操作，仍处于 "/a/b/c"

7. ".." 返回上级目录，最终为 "/a/b"

使用一个栈来解决问题。遇到‘..‘弹栈，遇到‘.‘不操作，其他情况下压栈。

代码一：

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

代码二：

利用python的字符串处理能力。

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

 


联系
我们