Leetcode 之Simplify Path @ python
Given an absolute path for a file (Unix-style), simplify it.
For example,
path = "/home/"
, => "/home"
path = "/a/./b/../../c/"
, => "/c"
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 =
res = ''
while i< len(path):
end = i+
while end<len(path) and path[end] !="/":
end +=
sub = path[i+:end]
if len(sub)>:
if sub == "..":
if stack !=[]:
stack.pop()
elif sub != ".":
stack.append(sub)
i = end if stack == []:
return "/"
for i in stack:
res += "/"+i
return res
code 2:
class Solution:
def simplifyPath(self,path):
path = path.split('/')
res = '/'
for i in path:
if i == '..':
if res != '/':
res = '/'.join(res.split('/')[:-1])
if res =='': res = '/'
elif i != '.' and i != '':
res += '/' +i if res != '/' else i
return res
转自(参考):
1. http://www.cnblogs.com/zuoyuan/p/3777289.html
2. http://blog.csdn.net/linhuanmars/article/details/23972563
@ JAVA 版本
public String simplifyPath(String path) {
if(path == null || path.length()==0)
{
return "";
}
LinkedList<String> stack = new LinkedList<String>();
StringBuilder res = new StringBuilder();
int i=0; while(i<path.length())
{
int index = i;
StringBuilder temp = new StringBuilder();
while(i<path.length() && path.charAt(i)!='/')
{
temp.append(path.charAt(i));
i++;
}
if(index!=i)
{
String str = temp.toString();
if(str.equals(".."))
{
if(!stack.isEmpty())
stack.pop();
}
else if(!str.equals("."))
{
stack.push(str);
}
}
i++;
}
if(!stack.isEmpty())
{
String[] strs = stack.toArray(new String[stack.size()]);
for(int j=strs.length-1;j>=0;j--)
{
res.append("/"+strs[j]);
}
}
if(res.length()==0)
return "/";
return res.toString();
}
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