Digit Wood

Write a class called MyRegex which will contain a string pattern. You need to write a regular expression and assign it to the pattern such that it can be used to validate an IP address. Use the following definition of an IP address:

IP address is a string in the form "A.B.C.D", where the value of A, B, C, and D may range from 0 to 255. Leading zeros are allowed. The length of A, B, C, or D can't be greater than 3.

Some valid IP address:

000.12.12.034
121.234.12.12
23.45.12.56

Some invalid IP address:

000.12.234.23.23
666.666.23.23
.213.123.23.32
23.45.22.32.
I.Am.not.an.ip

In this problem you will be provided strings containing any combination of ASCII characters. You have to write a regular expression to find the valid IPs.

Just write the MyRegex class which contains a String pattern . The string should contain the correct regular expression.

(MyRegex class MUST NOT be public)

Sample Input

000.12.12.034
121.234.12.12
23.45.12.56
00.12.123.123123.123
122.23
Hello.IP

Sample Output

true
true
true
false
false
false

SOLUTION: -1

import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.Scanner;

class Solution{

    public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        while(in.hasNext()){
            String IP = in.next();
            System.out.println(IP.matches(new MyRegex().pattern));
        }

    }
}
class MyRegex{

public static String pattern = "\\b(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\b";
}

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