From: Joel VanderWerf <joelvanderwerf@...>
Date: 2010-09-21T03:17:51+09:00
Subject: Re: An elegant way...

On 09/20/2010 11:06 AM, Joel VanderWerf wrote:
 > ...

Refactoring that:

a = [ '1', '2', '3', '4', '6', '7', '9', 'S1', 'S2', "10", "11", "12" ]
#a = [ '1', '2', 'S3', 'S4', 'S5', 'O6' ]

module Enumerable
   def each_run(cond)
     run = nil
     last = nil

     each_cons 2 do |prev, s|
       if cond[prev, s]
         run ||= [prev]
         run << s
       else
         yield run || prev
         run = nil
       end
       last = s
     end

     yield run || [last]

     self
   end

end

result = []

a.each_run(proc {|prev, s| prev.succ == s}) do |run|
   if run.size > 1
     result << "#{run.first}-#{run.last}"
   else
     result << run.first
   end
end

p result