[LeetCode] Simplify Path

[Problem]

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".

[Solution]

class Solution {
public:
    // split path by '/'
    vector split(string path){
        int i = 0;
        string str = "";
        vector res;
        while(i < path.length()){
            if(path[i] == '/'){
                if(str != ""){
                    res.push_back(str);
                    str = "";
                }
            }
            else{
                str += path[i];
            }
            i++;
        }
   
        // be careful the last path
        if(str != ""){
            res.push_back(str);
        }
        return res;
    }
   
    // simplify path
    string simplifyPath(string path) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        deque myQueue;
        myQueue.push_back("/");
   
        // split path
        vector dirs = split(path);
   
        // enqueue
        for(int i = 0; i < dirs.size(); ++i){
            if(dirs[i] == "."){
                continue;
            }
            else if(dirs[i] == ".."){
                if(myQueue.back() == "/" && myQueue.size() == 1){
                    continue;
                }
                else{
                    // pop dir
                    myQueue.pop_back();
   
                    // pop '/'
                    if(myQueue.size() > 1){
                        myQueue.pop_back();
                    }
                }
            }
            else{
                if(myQueue.back() != "/"){
                    myQueue.push_back("/");
                }
                myQueue.push_back(dirs[i]);
            }
        }
   
        // dequeue
        string res = "";
        while(!myQueue.empty()){
            res += myQueue.front();
            myQueue.pop_front();
        }
        return res;
    }
};

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