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 . The string should contain the correct regular expression.

(MyRegex class MUST NOT be public)


hackerrank java regex solution


Java Regex problem solution | HackerRank

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));
        }

    }
}

//YOU SHOULD ONLY SUBMIT THE FOLLOWING BLOCK
class MyRegex
{
         String pattern = 
            "^([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
            "([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
            "([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
            "([01]?\\d\\d?|2[0-4]\\d|25[0-5])$";

}