LeetCode 71. Simplify Path
一、 問題描述:
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 multipleslashes '/' together, such as "/home//foo/".
In this case, you should ignore redundant slashes and return "/home/foo".
二、 問題分析:
給定Unix-style的絕對路徑,要求簡化路徑,找出沒有迴路或多餘路徑的最簡化路徑。
首先,先了解下Unix-style的絕對路徑格式:
linux的絕對路徑是指從根目錄說起的. 例如 /dev/somedir/...
有5個相對路徑的表示方法:
1. 當前目錄 /.
2. 父目錄 /..
3. 指定目錄 /filename
4. 某使用者的根目錄 ~user
5. 自己的根目錄 ~
注意:後面兩種表示方法本題不會出現
例如,給定 "/a/./b/../../c/":
1、 因為/.表示當前目錄,所以忽略這一路徑,路徑縮減為"/a/b/../../c/";
2、 /..表示父目錄,需退回前一個路徑,路徑縮減為"/a/../c/";
3、 所以最後輸出結果為"/c";
另外,需特殊考慮:
1、 當 path = "/../"時,返回路徑應為"/";
2、 忽略重複的'/',例如:"/home//foo/",應返回 "/home/foo";
三、 演算法設計:
因為路徑屬於層次性結構,類似遞迴函式,在一層中指定新的一層,在新的一層中又重複上一層的操作。而又會出現回退一層的情況,這不難和棧(stack)這一資料結構聯想在一起:指定新一層等同於壓棧操作(push),回退一層等同於彈棧操作(pop),直至再無需要壓棧的元素後,路徑壓縮即完成。
演算法描述如下:
給定路徑path,定義空字串str,空列表l,遍歷path中的字元ch,進行以下操作:
1 如果ch不為‘/’,則令str = str + ch
2 如果ch為‘/’且str不為空串,則對str進行以下操作:
2.1 如果str為‘.’(當前目錄),則僅把str賦值為空字串
2.2 如果str為‘..’(父目錄),若l不為空,則刪除l中最後一個元素並把str賦值為空字串;若l為空,則僅把str賦值為空字串
2.3 其他情況時,均把str新增進l,並把str賦值為空字串
3 如果ch為path中最後一個字元,則判斷str是否為空,不為空則把str新增進l
最後獲得的列表l即為最短絕對路徑目錄,在每個目錄中間加上‘/’後輸出即為最終簡化後的絕對路徑
四、 程式實現:
class Solution:
def simplifyPath(self, path):
"""
:type path: str
:rtype: str
"""
path = list(path)
stack = []
n = len(path)
name = ''
for a in range(n):
if(path[a]=='/' and name!=''):
self.fun(stack, name)
name = ''
elif(path[a]!='/'):
name += path[a]
if(name!=''):
self.fun(stack, name)
s = ''
while(stack):
s = s + '/' + stack[0]
del stack[0]
if(s==''):
s = '/'
return s
def fun(self, stack, name):
if(name=='..'):
if(stack):
stack.pop()
elif(name=='.'):
return
else:
stack.append(name)
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