71. Simplify Path

Description

You are given an absolute path for a Unix-style file system, which always begins with a slash '/'. Your task is to transform this absolute path into its simplified canonical path.

The rules of a Unix-style file system are as follows:

• A single period '.' represents the current directory.
• A double period '..' represents the previous/parent directory.
• Multiple consecutive slashes such as '//' and '///' are treated as a single slash '/'.
• Any sequence of periods that does not match the rules above should be treated as a valid directory or file name. For example, '...' and '....' are valid directory or file names.

The simplified canonical path should follow these rules:

• The path must start with a single slash '/'.
• Directories within the path must be separated by exactly one slash '/'.
• The path must not end with a slash '/', unless it is the root directory.
• The path must not have any single or double periods ('.' and '..') used to denote current or parent directories.

Return the simplified canonical 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 (the parent directory).

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)
}

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