LeetCode: 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".

---

We can use a stack to help deal with the '/..' backwards case.

C++ Code:

/*
 * func: simplify_path
 * goal: simplify input absolute path
 * @param path: input path
 * return: simplified path
 */
string simplify_path(const string &path){
    string result = "";
    string dir = "";
    deque<string> path_helper;
    for(const char &ch : path){
        if(ch != '/'){
            dir += ch;
        }else{
            //If current char is '/' and previous char is also '/'
            //Skip this one
            if(dir == "/"){
                continue;
            }else if(dir == "/."){
                //If previous dir is '/.', nothing changed, continue to read next directory
                dir = "/";
            }else if(dir == "/.."){
                //If previous dir is '/..', pop one from stack if there exists
                if(!path_helper.empty()){
                    path_helper.pop_back();
                }
                dir = "/";
            }else{
                //If previous dir is not a special case, push back it to stack, and read the next one
                if(dir != "")
                    path_helper.emplace_back(dir);
                dir = "/";
            }
        }
    }
    //If the last one is '/..', pop if available
    if(dir == "/.."){
        if(!path_helper.empty()){
            path_helper.pop_back();
        }
    }else if(dir != "/" && dir != "/."){
        //If the last one is a normal one, push back it to the stack
        path_helper.emplace_back(dir);
    }
    
    //If there is nothing in the stack, return root directory
    if(path_helper.empty()){
        return "/";
    }
    
    auto it = path_helper.begin();
    while(it != path_helper.end()){
        result += *it++;
    }
    
    return result;
}

Python Code:

# func: simplify the input absolute path
# @param path: input path
# @return: simplified path
def simplify_path(path):
    dirs = path.split("/")
    stack = []

    for directory in dirs:
        if not directory or directory == '.':
            continue
        elif directory == '..':
            if stack:
                stack = stack[:-1]
        else:
            stack.append(directory)

    if not stack:
        return '/'
    else:
        result = ""
        for directory in stack:
            result += "/" + directory

        return result