LeetCode Simplify Path

Given an absolute path for a file (Unix-style), simplify it.

For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"

click to show corner cases.

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".

class Solution {

public:

    string simplifyPath(string path) {

        int len = path.length();

        if (len < 2) return "/";



        vector<string> spath;



        int p = -1, q = 0;



        // seperate the path into parts & simplify it with a stack

        while (q < len) {

            while (q < len && path[q] != '/') q++;

            if (p + 1 < q) {

                add_part_to_spath(path.substr(p + 1, q - p - 1), spath);

            }

            p = q;

            q++;

        }



        // build the final simplified path

        string res;

        for (int i=0; i<spath.size(); i++) {

            res.push_back('/');

            res.append(spath[i]);

        }

        if (res.length() == 0) res = "/";

        return res;



    }



    void add_part_to_spath(string part, vector<string>& spath) {

        if (part == ".") {

            // do nothing;

        } else if (part == "..") {

            // goto the parent path part

            if (!spath.empty()) spath.pop_back();

        } else {

            // enter a new part

            spath.push_back(part);

        }



    }

};

 也可以直接在原有字符串上操作，但还是这样简单明了一些

再写一次：

class Solution {

public:

    string simplifyPath(string path) {

        vector<string> tokens;

        

        int len = path.size();

        

        int last = 0;

        for (int i=0; i<len; i++) {

            char ch = path[i];

            if (ch == '/') {

                if (i - last > 0) {

                    push_token(path.substr(last, i - last), tokens);

                }

                last = i + 1;

            }

        }

        if (len - last > 0) {

            push_token(path.substr(last, len - last), tokens);

        }

        return token_path(tokens);

    }

    

    string token_path(vector<string>& tokens) {

        string path;

        for (string& token : tokens) {

            path = path + "/" + token;

        }

        if (tokens.empty()) {

            path = "/";

        }

        return path;

    }

    

    void push_token(string token, vector<string>& tokens) {

        if (token == ".") {

            // nothing

        } else if (token == "..") {

            if (!tokens.empty()) {

                tokens.pop_back();

            }

        } else {

            tokens.push_back(token);

        }

    }

};