71. Simplify Path

Description

Given an absolute path for a Unix-style file system, which begins with a slash '/', transform this path into its simplified canonical path.

In Unix-style file system context, a single period '.' signifies the current directory, a double period ".." denotes moving up one directory level, and multiple slashes such as "//" are interpreted as a single slash. In this problem, treat sequences of periods not covered by the previous rules (like "...") as valid names for files or directories.

The simplified canonical path should adhere to the following rules:

• It must start with a single slash '/'.
• Directories within the path should be separated by only one slash '/'.
• It should not end with a slash '/', unless it's the root directory.
• It should exclude any single or double periods used to denote current or parent directories.

Return the new path.

 

Example 1:

Input: path = "/home/"

Output: "/home"

Explanation:

The trailing slash should be removed.

 

Example 2:

Input: path = "/home//foo/"

Output: "/home/foo"

Explanation:

Multiple consecutive slashes are replaced by a single one.

Example 3:

Input: path = "/home/user/Documents/../Pictures"

Output: "/home/user/Pictures"

Explanation:

A double period ".." refers to the directory up a level.

Example 4:

Input: path = "/../"

Output: "/"

Explanation:

Going one level up from the root directory is not possible.

Example 5:

Input: path = "/.../a/../b/c/../d/./"

Output: "/.../b/d"

Explanation:

"..." is a valid name for a directory in this problem.

 

Constraints:

• 1 <= path.length <= 3000
• path consists of English letters, digits, period '.', slash '/' or '_'.
• path is a valid absolute Unix path.

Solutions

Solution 1: Stack

We first split the path into a number of substrings split by '/'. Then, we traverse each substring and perform the following operations based on the content of the substring:

• If the substring is empty or '.', no operation is performed because '.' represents the current directory.
• If the substring is '..', the top element of the stack is popped, because '..' represents the parent directory.
• If the substring is other strings, the substring is pushed into the stack, because the substring represents the subdirectory of the current directory.

Finally, we concatenate all the elements in the stack from the bottom to the top of the stack to form a string, which is the simplified canonical path.

The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the length of the path.

Python3JavaC++GoTypeScriptRustC#

class Solution:
    def simplifyPath(self, path: str) -> str:
        stk = []
        for s in path.split('/'):
            if not s or s == '.':
                continue
            if s == '..':
                if stk:
                    stk.pop()
            else:
                stk.append(s)
        return '/' + '/'.join(stk)

class Solution {
    public String simplifyPath(String path) {
        Deque<String> stk = new ArrayDeque<>();
        for (String s : path.split("/")) {
            if ("".equals(s) || ".".equals(s)) {
                continue;
            }
            if ("..".equals(s)) {
                stk.pollLast();
            } else {
                stk.offerLast(s);
            }
        }
        return "/" + String.join("/", stk);
    }
}

class Solution {
public:
    string simplifyPath(string path) {
        deque<string> stk;
        stringstream ss(path);
        string t;
        while (getline(ss, t, '/')) {
            if (t == "" || t == ".") {
                continue;
            }
            if (t == "..") {
                if (!stk.empty()) {
                    stk.pop_back();
                }
            } else {
                stk.push_back(t);
            }
        }
        if (stk.empty()) {
            return "/";
        }
        string ans;
        for (auto& s : stk) {
            ans += "/" + s;
        }
        return ans;
    }
};

func simplifyPath(path string) string {
    var stk []string
    for _, s := range strings.Split(path, "/") {
        if s == "" || s == "." {
            continue
        }
        if s == ".." {
            if len(stk) > 0 {
                stk = stk[0 : len(stk)-1]
            }
        } else {
            stk = append(stk, s)
        }
    }
    return "/" + strings.Join(stk, "/")
}

function simplifyPath(path: string): string {
    const stk: string[] = [];
    for (const s of path.split('/')) {
        if (s === '' || s === '.') {
            continue;
        }
        if (s === '..') {
            if (stk.length) {
                stk.pop();
            }
        } else {
            stk.push(s);
        }
    }
    return '/' + stk.join('/');
}

impl Solution {
    #[allow(dead_code)]
    pub fn simplify_path(path: String) -> String {
        let mut s: Vec<&str> = Vec::new();

        // Split the path
        let p_vec = path.split("/").collect::<Vec<&str>>();

        // Traverse the path vector
        for p in p_vec {
            match p {
                // Do nothing for "" or "."
                "" | "." => {
                    continue;
                }
                ".." => {
                    if !s.is_empty() {
                        s.pop();
                    }
                }
                _ => s.push(p),
            }
        }

        "/".to_string() + &s.join("/")
    }
}

public class Solution {
    public string SimplifyPath(string path) {
        var stk = new Stack<string>();
        foreach (var s in path.Split('/')) {
            if (s == "" || s == ".") {
                continue;
            }
            if (s == "..") {
                if (stk.Count > 0) {
                    stk.Pop();
                }
            } else {
                stk.Push(s);
            }
        }
        var sb = new StringBuilder();
        while (stk.Count > 0) {
            sb.Insert(0, "/" + stk.Pop());
        }
        return sb.Length == 0 ? "/" : sb.ToString();
    }
}

Solution 2

Go

func simplifyPath(path string) string {
    return filepath.Clean(path)
}

Comments