Code Golf: Numeric Ranges

Answer 1

Common Lisp, 442/206 chars

(defun d (l)
  (if l
      (let ((f (car l))
        (r (d (cdr l))))
      (if r
          (if (= (+ f 1) (caar r))
          (push `(,f ,(cadar r)) (cdr r))
          (push `(,f ,f) r))
          `((,f ,f))
          ))
      nil))

(defun p (l)
  (mapc #'(lambda (x)
          (if (= (car x) (cadr x))
          (format t "~a " (car x))
          (if (= (+ 1 (car x)) (cadr x))
              (format t "~a ~a " (car x) (cadr x))
              (format t "~a-~a " (car x) (cadr x)))))
      (d l)))

The "d" function rewrites the input list into a canonical form. For fun I did this entirely recursively. The "p" function formats the output to the equivalent of the reference implementation.

Answer 2

Ruby : 123 characters

def y(n) t=[];r=[];n.each_with_index do |x,i| t<<x;if(x.succ!=n[i+1]);r=((t.size>2)?r<<t[0]<<-t[-1]:r+t);t=[];end;end;r;end

More Readable

def y(n) 
t=[];r=[];
n.each_with_index do |x,i|
 t << x
 if (x.succ != n[i+1])
    r = ((t.size > 2) ? r << t[0] << -t[-1] : r+t)  
    t=[]
 end
 end
 r
end

And execute like

 > n=[1, 2, 3, 4, 7, 8, 10, 12, 13, 14, 15]
 > y n
 => [1, -4, 7, 8, 10, 12, -15]