Restore IP Addresses | Leetcode

Given a string containing only digits, restore it by returning all possible valid IP address combinations.

For example:
Given "25525511135",
return ["255.255.11.135", "255.255.111.35"]. (Order does not matter)

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Solution: Use a recursive combinatorial algorithm. Whenever the last added character was not a dot, add a dot. Whenever the last added character was zero, and the current number is zero, then return because we cannot have anything after a leading zero. For numbers, add it only if it will keep the current number less than or equal to 255.

public class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> list = new ArrayList<>();
        StringBuilder sb = new StringBuilder();
        restore(list, sb, s, 0, 0, 0);
        return list;
    }
    
    private void restore(List<String> list, StringBuilder sb, String s, int index, int word, int curNum) {
        if (index == s.length() && word == 3 && sb.charAt(sb.length() - 1) != '.') {
            list.add(sb.toString());
        }
        if (index >= s.length() || word > 3) {
            return;
        }

        if (sb.length() > 0) {
            int prev = sb.charAt(sb.length() - 1);
            // Add dot whenever possible.
            if (prev != '.') {
                sb.append('.');
                restore(list, sb, s, index, word + 1, 0);
                sb.setLength(sb.length() - 1);
            }
            // If last digit was already zero, then cannot double-up on zeros.
            if (curNum == 0 && prev == '0') {
                return;
            }
        }

        // Add next digit to current number if within range.
        int ch = s.charAt(index);
        curNum = (curNum * 10) + (ch - '0');
        if (curNum <= 255) {
            sb.append((char) ch);
            restore(list, sb, s, index + 1, word, curNum);
            sb.setLength(sb.length() - 1);
        }
    }
}