0093. Restore IP Addresses
https://leetcode.com/problems/restore-ip-addresses
Description
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.
ac
class Solution {
public List<String> restoreIpAddresses(String s) {
List<String> res = new ArrayList<>();
// Edge cases
if (s == null || s.length() < 4) {
return res;
}
restore(0, s, new ArrayList<>(), res);
return res;
}
private void restore(int start, String s, List<Integer> note, List<String> res) {
if (note.size() == 4) {
if (start == s.length()) {
res.add(note.stream().map(String::valueOf).collect(Collectors.joining(".")));
}
return;
}
for (int i = 1; i < 4 && start + i <= s.length(); i++) {
String tmp = s.substring(start, start + i);
if (tmp.startsWith("0") && tmp.length() > 1) continue;
int num = Integer.parseInt(tmp);
if (num > 255) continue;
note.add(num);
restore(start + i, s, note, res);
note.remove(note.size() - 1);
}
}
}
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