Skip to content

468. Validate IP Address

Description

Given a string queryIP, return "IPv4" if IP is a valid IPv4 address, "IPv6" if IP is a valid IPv6 address or "Neither" if IP is not a correct IP of any type.

A valid IPv4 address is an IP in the form "x1.x2.x3.x4" where 0 <= xi <= 255 and xi cannot contain leading zeros. For example, "192.168.1.1" and "192.168.1.0" are valid IPv4 addresses while "192.168.01.1", "192.168.1.00", and "192.168@1.1" are invalid IPv4 addresses.

A valid IPv6 address is an IP in the form "x1:x2:x3:x4:x5:x6:x7:x8" where:

  • 1 <= xi.length <= 4
  • xi is a hexadecimal string which may contain digits, lowercase English letter ('a' to 'f') and upper-case English letters ('A' to 'F').
  • Leading zeros are allowed in xi.

For example, "2001:0db8:85a3:0000:0000:8a2e:0370:7334" and "2001:db8:85a3:0:0:8A2E:0370:7334" are valid IPv6 addresses, while "2001:0db8:85a3::8A2E:037j:7334" and "02001:0db8:85a3:0000:0000:8a2e:0370:7334" are invalid IPv6 addresses.

 

Example 1:

Input: queryIP = "172.16.254.1"
Output: "IPv4"
Explanation: This is a valid IPv4 address, return "IPv4".

Example 2:

Input: queryIP = "2001:0db8:85a3:0:0:8A2E:0370:7334"
Output: "IPv6"
Explanation: This is a valid IPv6 address, return "IPv6".

Example 3:

Input: queryIP = "256.256.256.256"
Output: "Neither"
Explanation: This is neither a IPv4 address nor a IPv6 address.

 

Constraints:

  • queryIP consists only of English letters, digits and the characters '.' and ':'.

Solutions

Solution 1: Simulation

We can define two functions isIPv4 and isIPv6 to determine whether a string is a valid IPv4 address and IPv6 address.

The implementation of the function isIPv4 is as follows:

  1. We first check if the string s ends with .. If so, s is not a valid IPv4 address, and we directly return false.
  2. Then we split the string s by . into a string array ss. If the length of ss is not 4, s is not a valid IPv4 address, and we directly return false.
  3. For each string t in the array ss, we check:
    • If the length of t is greater than 1 and the first character of t is 0, t is not a valid IPv4 address, and we directly return false.
    • If t is not a number or t is not in the range of 0 to 255, t is not a valid IPv4 address, and we directly return false.
  4. If none of the above conditions are met, s is a valid IPv4 address, and we return true.

The implementation of the function isIPv6 is as follows:

  1. We first check if the string s ends with :. If so, s is not a valid IPv6 address, and we directly return false.
  2. Then we split the string s by : into a string array ss. If the length of ss is not 8, s is not a valid IPv6 address, and we directly return false.
  3. For each string t in the array ss, we check:
    • If the length of t is less than 1 or greater than 4, t is not a valid IPv6 address, and we directly return false.
    • If the characters in t are not all between 0 and 9 and a and f (case insensitive), t is not a valid IPv6 address, and we directly return false.
  4. If none of the above conditions are met, s is a valid IPv6 address, and we return true.

Finally, we call the isIPv4 and isIPv6 functions to determine if queryIP is a valid IPv4 address or IPv6 address. If it is neither, we return Neither.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string queryIP.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
class Solution:
    def validIPAddress(self, queryIP: str) -> str:
        def is_ipv4(s: str) -> bool:
            ss = s.split(".")
            if len(ss) != 4:
                return False
            for t in ss:
                if len(t) > 1 and t[0] == "0":
                    return False
                if not t.isdigit() or not 0 <= int(t) <= 255:
                    return False
            return True

        def is_ipv6(s: str) -> bool:
            ss = s.split(":")
            if len(ss) != 8:
                return False
            for t in ss:
                if not 1 <= len(t) <= 4:
                    return False
                if not all(c in "0123456789abcdefABCDEF" for c in t):
                    return False
            return True

        if is_ipv4(queryIP):
            return "IPv4"
        if is_ipv6(queryIP):
            return "IPv6"
        return "Neither"
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
class Solution {
    public String validIPAddress(String queryIP) {
        if (isIPv4(queryIP)) {
            return "IPv4";
        }
        if (isIPv6(queryIP)) {
            return "IPv6";
        }
        return "Neither";
    }

    private boolean isIPv4(String s) {
        if (s.endsWith(".")) {
            return false;
        }
        String[] ss = s.split("\\.");
        if (ss.length != 4) {
            return false;
        }
        for (String t : ss) {
            if (t.length() == 0 || t.length() > 1 && t.charAt(0) == '0') {
                return false;
            }
            int x = convert(t);
            if (x < 0 || x > 255) {
                return false;
            }
        }
        return true;
    }

    private boolean isIPv6(String s) {
        if (s.endsWith(":")) {
            return false;
        }
        String[] ss = s.split(":");
        if (ss.length != 8) {
            return false;
        }
        for (String t : ss) {
            if (t.length() < 1 || t.length