Regex to match four repeated letters in a string using a Java pattern

Question

I want to match something like aaaa, aaaad, adjjjjk. Something like ([a-z])\1+ was used to match the repeated characters, but I am not able to figure this out for four letters.

Answer 1

You want to match a single character and then that character repeated three more times:

([a-z])\1{3}

Note: In Java you need to escape the backslashes inside your regular expressions.

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Update: The reason why it isn't doing what you want is because you are using the method matches which requires that the string exactly matches the regular expression, not just that it contains the regular expression. To check for containment you should instead use the Matcher class. Here is some example code:

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

class Program
{
    public static void main(String[] args)
    {
        Pattern pattern = Pattern.compile("([a-z])\\1{3}");
        Matcher matcher = pattern.matcher("asdffffffasdf");
        System.out.println(matcher.find());
    }
}

Result:

true

Answer 2

Not knowing about the finite repetition syntax, your own problem solving skill should lead you to this:

([a-z])\1\1\1

Obviously it's not pretty, but:

• It works
• It exercises your own problem solving skill
• It may lead you to deeper understanding of concepts
  • In this case, knowing the desugared form of the finite repetition syntax

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I have a concern:

• "ffffffff".matches("([a-z])\\1{3,}") = true
• "fffffasdf".matches("([a-z])\\1{3,}") = false
• "asdffffffasdf".matches("([a-z])\\1{3,}") = false

What can I do for the bottom two?

The problem is that in Java, matches need to match the whole string; it is as if the pattern is surrounded by ^ and $.

Unfortunately there is no String.containsPattern(String regex), but you can always use this trick of surrounding the pattern with .*:

"asdfffffffffasf".matches(".*([a-z])\\1{3,}.*") // true!
//                         ^^              ^^