In this post, we are going to solve the Simplify Path Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Problem
Given a string path
, which is an absolute path (starting with a slash '/'
) to a file or directory in a Unix-style file system, convert it to the simplified canonical path.
In a Unix-style file system, a period '.'
refers to the current directory, a double period '..'
refers to the directory up a level, and any multiple consecutive slashes (i.e. '//'
) are treated as a single slash '/'
. For this problem, any other format of periods such as '...'
are treated as file/directory names.
The canonical path should have the following format:
- The path starts with a single slash
'/'
. - Any two directories are separated by a single slash
'/'
. - The path does not end with a trailing
'/'
. - The path only contains the directories on the path from the root directory to the target file or directory (i.e., no period
'.'
or double period'..'
)
Return the simplified canonical path.
Example 1:
Input: path = "/home/" Output: "/home" Explanation: Note that there is no trailing slash after the last directory name.
Example 2:
Input: path = "/../" Output: "/" Explanation: Going one level up from the root directory is a no-op, as the root level is the highest level you can go.
Example 3:
Input: path = "/home//foo/" Output: "/home/foo" Explanation: In the canonical path, multiple consecutive slashes are replaced by a single one.
Constraints:
1 <= path.length <= 3000
path
consists of English letters, digits, period'.'
, slash'/'
or'_'
.path
is a valid absolute Unix path.
Now, let’s see the leetcode solution of Simplify Path Leetcode Solution.
Simplify Path Leetcode Solution in Python
class Solution: def simplifyPath(self, path: str) -> str: stack = [] for str in path.split('/'): if str in ('', '.'): continue if str == '..': if stack: stack.pop() else: stack.append(str) return '/' + '/'.join(stack)
Simplify Path Leetcode Solution in CPP
class Solution { public: string simplifyPath(string path) { string ans; istringstream iss(path); vector<string> stack; for (string dir; getline(iss, dir, '/');) { if (dir.empty() || dir == ".") continue; if (dir == "..") { if (!stack.empty()) stack.pop_back(); } else { stack.push_back(dir); } } for (const string& s : stack) ans += "/" + s; return ans.empty() ? "/" : ans; } };
Simplify Path Leetcode Solution in Java
class Solution { public String simplifyPath(String path) { final String[] dirs = path.split("/"); Stack<String> stack = new Stack<>(); for (final String dir : dirs) { if (dir.isEmpty() || dir.equals(".")) continue; if (dir.equals("..")) { if (!stack.isEmpty()) stack.pop(); } else { stack.push(dir); } } return "/" + String.join("/", stack); } }
Note: This problem Simplify Path is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.