Simplify Path

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”.
这道题的要求是简化一个Unix风格下的文件的绝对路径。

字符串处理，由于”..”是返回上级目录（如果是根目录则不处理），因此可以考虑用栈记录路径名，以便于处理。需要注意几个细节：
使用一个vector来维护。

重复连续出现的’/’，只按1个处理，即跳过重复连续出现的’/’；
如果路径名是”.”，则不处理；
如果路径名是”..”，则需要删除，如果栈为空，则不做处理；
如果路径名为其他字符串，加入vector。

class Solution {public:    string simplifyPath(string path) {        vector<string> str;        string tmp = "";        for (int i = 0; i < path.size(); i++)        {            while(path[i] == '/' &&i < path.size())                i++;            string tmp = "";            while(path[i] != '/' && i < path.size()){                tmp += path[i];                i++;            }            if(tmp == ".." && !str.empty())                str.pop_back();            else if(tmp != ".." && tmp != "/" && tmp != "."&& tmp!="")                str.push_back(tmp);        }        if(str.empty())            return "";        string res = "";        for (int i = 0; i < str.size(); i++)        {            res = res + "/" +str[i];        }        return res;    }};