71. Simplify Path ---------M !!!!

Q:

You are given an absolute path for a Unix-style file system, which always begins with a slash '/'. Your task is to transform this absolute path into its simplified canonical path.

The rules of a Unix-style file system are as follows:

• A single period '.' represents the current directory.
• A double period '..' represents the previous/parent directory.
• Multiple consecutive slashes such as '//' and '///' are treated as a single slash '/'.
• Any sequence of periods that does not match the rules above should be treated as a valid directory or file name. For example, '...' and '....' are valid directory or file names.

The simplified canonical path should follow these rules:

• The path must start with a single slash '/'.
• Directories within the path must be separated by exactly one slash '/'.
• The path must not end with a slash '/', unless it is the root directory.
• The path must not have any single or double periods ('.' and '..') used to denote current or parent directories.

Return the simplified canonical path.

 

Example 1:

Input: path = "/home/"

Output: "/home"

Explanation:

The trailing slash should be removed.

Example 2:

Input: path = "/home//foo/"

Output: "/home/foo"

Explanation:

Multiple consecutive slashes are replaced by a single one.

Example 3:

Input: path = "/home/user/Documents/../Pictures"

Output: "/home/user/Pictures"

Explanation:

A double period ".." refers to the directory up a level (the parent directory).

Example 4:

Input: path = "/../"

Output: "/"

Explanation:

Going one level up from the root directory is not possible.

Example 5:

Input: path = "/.../a/../b/c/../d/./"

Output: "/.../b/d"

Explanation:

"..." is a valid name for a directory in this problem.

 

Constraints:

• 1 <= path.length <= 3000
• path consists of English letters, digits, period '.', slash '/' or '_'.
• path is a valid absolute Unix path.

A:


------------------第二遍， 学乖了，直接用了个split函数-----------

class Solution {
public:
    string simplifyPath(string path) {
        vector<string> V;
        int start = 0, end = 0;
        while(start < path.size()){
            if(path[start] == '/' ){
                start++;
                continue;
            }else{
                end = start;
                while(end < path.size() && path[end] != '/'){
                    end++;
                }
                string newName = path.substr(start, end-start);
                if(newName == ".."){
                    if( ! V.empty())
                        V.pop_back();
                }else if(newName != "."){
                    V.push_back(newName);
                }
                start = end;
            }
        }
        string res;
        for(auto s:V)
            res += "/" + s;
        return res.size() == 0? "/" : res;
    }
};

Mistakes：
1： 没有考虑，当buf为空的时候，我们仍然要输出一个"/"
2:    split（）函数不太明白。   对于“/..."这个来讲，分出的是两个string { ""，"..."}
        因此，我们需要处理curPath为空的情况。

 --------------1st pass-------------------
这道题就不是考 难度的，是考 细心！！！！！！！！！！！！！！！！！

Mistakes:
 1:取 substring的时候，忘记把 lastSlashPos 的位置+1了。  导致，总是包括了两个//
 2:  Did you consider the case where path = "/../"?
       In this case, you should return "/".
 当上述情况时，我们不能太简单的删除last element of ArrayList, 而应该检查是否是到了根节点了（ArrayList.isEmpty()。）
3: 当 ArrayList为空时， 还是要输出一个  "/" 的， 哎， 竟然这样SB的错误啊~~~~~~~~
4: 没有考虑，当最后的是 “."   ".."的情况，╮(╯▽╰)╭
   太SB得直接append到ArrayList上了。
5： path.substring(lastSlashPos+1);    这里也是要加一的。 ╮(╯▽╰)╭
全线SB啊，Tiger

 Learned: