Java Regex HackerRank Solution
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
Code Solution:
#Java Regex HackerRank Solution
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{
String zeroTo255
= "(\\d{1,2}|(0|1)\\"
+ "d{2}|2[0-4]\\d|25[0-5])";
String pattern
= zeroTo255 + "\\."
+ zeroTo255 + "\\."
+ zeroTo255 + "\\."
+ zeroTo255;
}
#Java Regex HackerRank Solution
Disclaimer: This problem is originally created and published by HackerRank, we only provide solutions to this problem. Hence, doesn’t guarantee the truthfulness of the problem. This is only for information purposes.
