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 wherePath="/../"?
  In this case, you shoshould return"/".
• Another corner case is the path might contain multiple slashes‘/‘Together, such"/home//foo/".
  In this case, you shoshould ignore redundant slashes and return"/home/foo".

The question is not difficult, mainly considering some special situations.

ForPath="/a/./b/../../c/", =>"/C", simulate

Split the string according to '/' to obtain [A,., B,..., C].

First, go to directory A. Note that '.' indicates the current directory, and ".." indicates the previous directory.

Then it reaches '.', still in the current directory,/

Then it reaches 'B', Which is/a/B.

Then, it reaches '..'. This is returned to the parent directory, and it becomes/.

Then, it reaches '..' and continues to return to the parent directory. Then it becomes/

Then it reaches 'C', and the sub-directory is reached to/C.

class Solution {public:    vector<string> split(string& path, char ch){        int index = 0;        vector<string> res;        while(index < path.length()){            while(index < path.length() && path[index] == ‘/‘) index++;            if(index >= path.length()) break;            int start=index, len = 0;            while(index < path.length() && path[index]!=‘/‘) {index++;len++;}            res.push_back(path.substr(start,len));        }        return res;    }        string simplifyPath(string path) {        vector<string> a = split(path,‘/‘);        vector<string> file;        for(int i = 0 ; i < a.size(); ++ i){            if(a[i] == ".." ){                if(!file.empty()) file.pop_back();            }            else if(a[i]!=".") file.push_back(a[i]);        }        string res="";        if(file.empty()) res ="/";        else{            for(int i = 0 ; i < file.size(); ++ i) res+="/"+file[i];        }        return res;    }};