Java Regex: Grouping consecutive 1 or 0 in a binary string

Question

I want to capture all the consecutive groups in a binary string

1000011100001100111100001

should give me

I have made ([1?|0?]+) regex in my java application to group the consequential 1 or 0 in the string like 10000111000011. But when I run it in my code, there is nothing in the console printed:

String name ="10000111000011";
    regex("(\\[1?|0?]+)" ,name);

    public static void regex(String regex, String searchedString) {

        Pattern pattern = Pattern.compile(regex);
        Matcher regexMatcher = pattern.matcher(searchedString);
        while (regexMatcher.find()) 
            if (regexMatcher.group().length() > 0)
                System.out.println(regexMatcher.group());
    }

To avoid syntax error in the runtime of regex, I have changed the ([1?|0?]+) to the (\\[1?|0?]+)

Why there is no group based on regex?

i edited my question, I know. it just consider all as one group. But u shouldn't downvote me because of it :( — Sal-laS
– Sal-laS, Commented Jan 18, 2017 at 11:33
Next time please provide sample input and output from the start :) — Quota
– Quota, Commented Jan 18, 2017 at 11:43
Yes, sure. I just thought, the tester site accepts it, but i didn't noticed it accepted it as only one group — Sal-laS
– Sal-laS, Commented Jan 18, 2017 at 11:48

SamWhan · Accepted Answer · 2017-01-18 12:12:24Z

2

First - just as an explanation - your regex defines a character class ([ ... ]) that matches any of the characters 1, ?, | or 0 one or more times (+). I think you mean to have ( ... ) in it, among other things, which would make the | an alternation lazy matching a 0 or a 1. But that's not either what you want (I think ;).

Now, the solution might be this:

([01])\1*

which matches a 0 or a 1, and captures it. Then it matches any number of the same digit (\1 is a back reference to what ever is captured in the first capture group - in this case the 0 or the 1) any number of times.

Check it out at ideone.

edited Jan 18, 2017 at 12:12

answered Jan 18, 2017 at 11:07

SamWhan

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10 Comments

Sal-laS Over a year ago

Your regex also, does not print anything (\[01])\1

SamWhan Over a year ago

Check the example added.

SamWhan Over a year ago

Added explanation to answer.

SamWhan Over a year ago

@Quota Correct, it captures the first digit. But if you check the OP's example, the full match is used - regexMatcher.group() (No number as parameter to group gives the complete match).

SamWhan Over a year ago

@Quota That may be true - it might be simpler to understand. And in that case you may even simplify it further by removing the capturing group - 0+|1+ which makes it (much) more efficient. The one I've given can easily be modified for other applications though (like matching ranges of letters (\w)\1*). But that's not really in the scope, so...

|

Mustofa Rizwan · Accepted Answer · 2017-01-18 11:03:27Z

1

You can try this:

(1+|0+)

Explanation

Sample Code:

    final String regex = "(1+|0+)";
    final String string = "10000111000011\n"
            + "11001111110011";

    final Pattern pattern = Pattern.compile(regex, Pattern.CASE_INSENSITIVE | Pattern.UNICODE_CASE | Pattern.MULTILINE);
    final Matcher matcher = pattern.matcher(string);

    while (matcher.find()) {

                System.out.println("Group " + 1 + ": " + matcher.group(1));


    }

edited Jan 18, 2017 at 11:03

answered Jan 18, 2017 at 10:58

Mustofa Rizwan

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