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".

題目不難，主要考慮一些特殊情況。

對於path = "/a/./b/../../c/", => "/c"，模擬一下

先按照'/'對字串進行分割，得到 [a, . , b, .. , .. , c]

首先進入目錄a，注意 '.' 代表當前目錄 ，".."代表上一個目錄

然後到達'.'，還是在當前目錄，/a

然後到達'b'，這為/a/b

然後到達'..'，這是回到父目錄，則變為/a

然後到達'..'，繼續回到父目錄，則變為/

然後到達'c'，則達到子目錄，變為/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;
    }
};