Given an absolute path for a file (Unix-style), simplify it.
For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"
path = "/a/../../b/../c//.//", => "/c"
path = "/a//b////c/d//././/..", => "/a/b/c"
In a UNIX-style file system, a period ('.') refers to the current directory, so it can be ignored in a simplified path. Additionally, a double period ("..") moves up a directory, so it cancels out whatever the last directory was. For more information, look here: https://en.wikipedia.org/wiki/Path_(computing)#Unix_style
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".
解法：遇到／直接跳过，只看／／之间 的东西，放到栈里面，小心边界条件，最后不加／的那一种("/aaaa//bbbb////cdddd/d//././/..")
class Solution {
public:
string simplifyPath(string path)
{
string ans="";
vector<string> ans_reverse;
string temp="";
stack<string> sta;
for(auto x:path)//遍历所有字符
{
if(x=='/')//记住改成转义意符
{
if(temp.size()==0)// 缓存字符串啥也没有 没用的“\”
{
continue;
}
else//temp有东西 判断temp不同三种情况 最后temp清空
{
if(temp==".")
{
}
else if(temp=="..")
{
if(!sta.empty())sta.pop();
}
else
{
sta.push(temp);
// std::cout<<temp<<std::endl;
}
temp="";
}
}
else
{
temp=temp+x;
//std::cout<<temp<<std::endl;
}
}
if(temp!="")
{
if(temp==".")
{
}
else if(temp=="..")
{
if(!sta.empty())sta.pop();
}
else
{
sta.push(temp);
// std::cout<<temp<<std::endl;
}
}
while (!sta.empty())
{
ans_reverse.push_back(sta.top());
sta.pop();
}
if(ans_reverse.size()==0) return"/";
for(int i=ans_reverse.size()-1;i>=0;i--)
{
ans=ans+"/"+ans_reverse[i];
//std::cout<<ans_reverse[i]<<std::endl;
}
std::cout<<ans<<std::endl;
return ans;
}
};