93, Restore IP Addresses

I Problem

A valid IP address consists of exactly four integers separated by single dots. Each integer is between 0 and 255 (inclusive) and cannot have leading zeros.

• For example, "0.1.2.201" and "192.168.1.1" are valid IP addresses, but "0.011.255.245", "192.168.1.312" and "192.168@1.1" are invalid IP addresses.

Given a string s containing only digits, return all possible valid IP addresses that can be formed by inserting dots into s. You are not allowed to reorder or remove any digits in s. You may return the valid IP addresses in any order.

Example 1
Input: s = "25525511135"
Output: ["255.255.11.135", "255.255.111.35"]

Example 2
Input: s = "0000"
Output: ["0.0.0.0"]

Example 3
Input: s = "101023"
Output: ["1.0.10.23", "1.0.102.3", "10.1.0.23", "10.10.2.3", "101.0.2.3"]

Constraints

• 1 <= s.length <= 20
• s consists of digits only.

Related Topics

• String
• Backtracking

II Solution

Approach 1: Brute Force

pub fn restore_ip_addresses(s: String) -> Vec<String> {
    const DFS: for<'a> fn(usize, &'a str, &mut Vec<&'a str>, &mut Vec<String>) =
        |i, s, address, res| {
            if address.len() == 4 {
                res.push(address.join("."));
                return;
            }

            let start = if address.len() != 3 { i + 1 } else { s.len() };
            
            for j in start..=s.len() {
                let substr = &s[i..j];
                if substr.is_empty() {
                    break;
                }
                if substr.starts_with('0') && substr.len() > 1 {
                    break;
                }
                if substr.parse::<usize>().is_ok_and(|num| num > 255) {
                    break;
                }

                address.push(substr);
                DFS(j, s, address, res);
                address.pop();
            }
        };
    let mut res = vec![];

    DFS(0, &s, &mut vec![], &mut res);

    res
}

@FunctionalInterface
interface QuadrConsumer<A, B, C, D> {
    void accept(A a, B b, C c, D d);
}

QuadrConsumer<Integer, String, List<String>, List<String>> dfs =
        (i, s, address, res) -> {
            if (address.size() == 4) {
                res.add(String.join(".", address));
                return;
            }

            Integer start = i + 1;
            if (address.size() == 3) {
                start = s.length();
            }

            for (int j = start; j <= s.length(); j++) {
                String substr = s.substring(i, j);
                if (substr.isEmpty()) {
                    break;
                }
                if (substr.startsWith("0") && substr.length() > 1) {
                    break;
                }
                if (Long.parseUnsignedLong(substr) > 255) {
                    break;
                }

                address.addLast(substr);
                this.dfs.accept(j, s, address, res);
                address.removeLast();
            }
        };

public List<String> restoreIpAddresses(String s) {
    List<String> res = new ArrayList<>();

    this.dfs.accept(0, s, new ArrayList<>(), res);

    return res;
}