Leetcode Simplify Path

標籤：style   blog   color   strong   art   for   

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"，類比一下

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

首先進入目錄a，注意 ‘.‘ 代表目前的目錄 ，".."代表上一個目錄

然後到達‘.‘，還是在目前的目錄，/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;    }};