【LeetCode with Python】 Simplify Path

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原题页面：https://oj.leetcode.com/problems/simplify-path/

题目类型：

难度评价：★

本文地址：/article/1377468.html

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"

.

class Solution:
    # @param path, a string
    # @return a string
    def simplifyPath(self, path):
        stack = [ ]
        toks = path.split("/")
        stack.append(toks[0])
        toks = toks[1:]
        for tok in toks:
            if tok in ("", "."):
                continue
            elif ".." ==  tok:
                if len(stack) > 0 and "" != stack[len(stack) - 1]:   # root / should not be poped to distinguish abs path and rel path
                    stack.pop()
                #else:
                #    return ""         # /home/../../..
            else:
                stack.append(tok)
        len_stack = len(stack)
        if len_stack >1:
            return "/".join(stack) if "" == stack[0] else "/" + "/".join(stack)
        elif (1 == len_stack and "" == stack[0]) or 0 == len_stack:
            return "/"