Leetcode 71：Simplify Path

Given an absolute path for a file (Unix-style), simplify it. Or in other words, convert it to the canonical path.

In a UNIX-style file system, a period . refers to the current directory. Furthermore, a double period … moves the directory up a level. For more information, see: Absolute path vs relative path in Linux/Unix

Note that the returned canonical path must always begin with a slash /, and there must be only a single slash / between two directory names. The last directory name (if it exists) must not end with a trailing /. Also, the canonical path must be the shortest string representing the absolute path.

Example 1:

Input: “/home/”
Output: “/home”
Explanation: Note that there is no trailing slash after the last directory name.
Example 2:

Input: “/…/”
Output: “/”
Explanation: Going one level up from the root directory is a no-op, as the root level is the highest level you can go.
Example 3:

Input: “/home//foo/”
Output: “/home/foo”
Explanation: In the canonical path, multiple consecutive slashes are replaced by a single one.
Example 4:

Input: “/a/./b/…/…/c/”
Output: “/c”
Example 5:

Input: “/a/…/…/b/…/c//.//”
Output: “/c”
Example 6:

Input: “/a//b////c/d//././/…”
Output: “/a/b/c”

这道题要求我们去简化路径，用过linux的估计都知道..表示上一个目录，.标识当前目录，然后做这道题的话，就是按/将字符串分割，遇到..就去掉之前的字符串，遇到.就忽略。最后再补上/。

C++
string simplifyPath(string path) {
        vector<string> strs;
        int len = path.size();
        int i = 0;
        string s = "";
        while(i < len)
        {
            while(i<len&&path[i] == '/')
            {
                ++i;
            }
            if(i>=len)break;
            s = "";
            while(i<len&&path[i] != '/')
            {
                s.push_back(path[i]);
                ++i;
            }
            if(s == "..")
            {
                if(!strs.empty())
                    strs.pop_back();
            }
            else if(s != ".")
                strs.push_back(s);
        }
        if(strs.empty())return "/";
        string res = "";
        for(int i = 0;i < strs.size();++i)
        {
            res = res + "/" + strs[i];
        }
        return res;
    }