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【LeetCode】71. Simplify Path 解题报告(Python)


【LeetCode】71. Simplify Path 解题报告(Python)

标签(空格分隔): LeetCode

id: fuxuemingzhu

个人博客: ​​​http://fuxuemingzhu.me/​​​

题目地址:​​https://leetcode.com/problems/simplify-path/description/​​

题目描述:

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linux路径。

解题方法

看到这种来来回回,增增删删的题,一般都想到用栈。

我们把字符串按照/分割之后就得到了每个文件的目录,然后判断路径是添加还是向上层进行返回。这个题很简单了。

有一个地方犯了小错误,不能写成if dir == ‘..’ and stack: stack.pop()。这样的话如果栈是空的,就把..进栈了。

class Solution(object):
def simplifyPath(self, path):
"""
:type path: str
:rtype: str
"""
stack = list()
dirs = path.split('/')
for dir in dirs:
if not dir or dir == '.':
continue
if dir == '..':
if stack:
stack.pop()
else:
stack.append(dir)
return '/' + '/'.join(stack)

日期

2018 年 6 月 26 日 ———— 早睡早起



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