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Java Regex Hacker Rank Java 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 . 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
falseSolution:-
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));
}
}
}
//Write your code here
class MyRegex {
public String pattern = "((\\d|\\d\\d|[0-1]\\d\\d|2[0-4][0-9]|25[0-5])\\.){3}(\\d|\\d\\d|[0-1]\\d\\d|2[0-4][0-9]|25[0-5])";
}
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