LeetCode-in-Java
93. Restore IP Addresses
Medium
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 = “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 <= 20sconsists of digits only.
Solution
import java.util.ArrayList;
import java.util.List;
public class Solution {
private static final int SEG_COUNT = 4;
private List<String> result = new ArrayList<>();
private int[] segments = new int[SEG_COUNT];
public List<String> restoreIpAddresses(String s) {
dfs(s, 0, 0);
return result;
}
public void dfs(String s, int segId, int segStart) {
// find 4 segments and get to last index
if (segId == SEG_COUNT) {
if (segStart == s.length()) {
StringBuilder addr = new StringBuilder();
for (int i = 0; i < SEG_COUNT; i++) {
addr.append(segments[i]);
if (i != SEG_COUNT - 1) {
addr.append('.');
}
}
result.add(addr.toString());
}
return;
}
// last index and no 4 segments
if (segStart == s.length()) {
return;
}
// start with a zero
if (s.charAt(segStart) == '0') {
segments[segId] = 0;
dfs(s, segId + 1, segStart + 1);
return;
}
int addr = 0;
for (int index = segStart; index < s.length(); index++) {
addr = addr * 10 + s.charAt(index) - '0';
if (addr >= 0 && addr <= 255) {
segments[segId] = addr;
dfs(s, segId + 1, index + 1);
} else {
break;
}
}
}
}