re = /\b(?!\d*(\d)\1)[10]+\b/m
str = 'no binary numbers here 3434.
Hey friend this is a 1.
Those are 1001, 1010, 1011, 1100, 1101
This is a long value 1010101010 and this one as well 1010101010101011
0 + 0 is a also a 0.'
# Print the match result
str.scan(re) do |match|
puts match.to_s
end
Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for Ruby, please visit: http://ruby-doc.org/core-2.2.0/Regexp.html