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Leetcode Simplify Path

2024-07-12 04:21:54   214人阅读

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".

题目不难，主要考虑一些特殊情况。

对于path = "/a/./b/../../c/", => "/c"，模拟一下

先按照‘/‘对字符串进行分割，得到 [a, . , b, .. , .. , c]

首先进入目录a，注意 ‘.‘ 代表当前目录 ，".."代表上一个目录

然后到达‘.‘，还是在当前目录，/a

然后到达‘b‘，这为/a/b

然后到达‘..‘，这是回到父目录，则变为/a

然后到达‘..‘，继续回到父目录，则变为/

然后到达‘c‘，则达到子目录，变为/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;    }};

 


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