[Leetcode]@python 71. Simplify Path

题目链接

https://leetcode.com/problems/simplify-path/

题目原文

Given an absolute path for a file (Unix-style), simplify it.

For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"

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".

题目大意

简化Unix系统的文件访问路径

解题思路

先根据‘/’对字符串进行切割，用栈对路径名进行存储，若遇到路径名，则入栈；若遇到“..”，则将栈顶元素pop出来；最后将栈内剩余的元素用“/”连接后输出

代码

class Solution(object):
    def simplifyPath(self, path):
        """
        :type path: str
        :rtype: str
        """
        stack = []
        i = 0
        ans = ''
        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:
            ans += '/' + i
        return ans