Given a string s containing only digits, return all possible valid IP addresses that can be obtained from s. You can return them in any order.
A valid IP address consists of exactly four integers, each integer is between 0 and 255, separated by single dots and cannot have leading zeros. For example, “0.1.2.201” and “192.168.1.1” are valid IP addresses and “0.011.255.245”, “192.168.1.312” and “192.168@1.1” are invalid IP addresses.
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 = "1111"
Output: ["1.1.1.1"]
Example 4:
Input: s = "010010"
Output: ["0.10.0.10","0.100.1.0"]
Example 5:
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:
- 0 <= s.length <= 3000
- s consists of digits only.
Solution
class Solution {
public List<String> restoreIpAddresses(String s) {
List<String> res = new ArrayList<>();
restore(s, "", 0, res);
return res;
}
void restore(String s, String ip, int index, List<String> res) {
if (s.length() == 0) {
if (index == 4) res.add(ip);
return;
}
if (index > 4) return;
String digit = s.substring(0, 1);
String octet = ip.length() == 0 ? digit: ip + "." + digit;
restore(s.substring(1, s.length()), octet, index + 1, res);
if (s.startsWith("0")) return;
if (s.length() > 1) {
String twoDigits = s.substring(0, 2);
octet = ip.length() == 0 ? twoDigits: ip + "." + twoDigits;
restore(s.substring(2, s.length()), octet, index + 1, res);
}
if (s.length() > 2) {
String threeDigits = s.substring(0, 3);
octet = ip.length() == 0 ? threeDigits: ip + "." + threeDigits;
if (Integer.valueOf(threeDigits) <= 255) {
restore(s.substring(3, s.length()), octet, index + 1, res);
}
}
}
}