How embarrassing. My first version of problem 9 (packing lists into sublists) had a bug. Here's a cleaner version which actually works. I know the bug in problem 9. Can you spot it?
sub pack (@array) returns Array {
my @unpacked = @array;
my (@list, @packed);
while @unpacked {
@list.push(@unpacked.shift) while !@list || @list[0] eq @unpac
+ked[0];
@packed.push([@list]);
@list = ();
}
return @packed;
}
pack(<a a a a b c c a a d e e e e>).perl.say;
And for the Lisp weenies who claim Lisp is so much better, here's one way to do it in Lisp (can you make it shorter?):
(defun pack (lista)
(if (eql lista nil)
nil
(cons (pega lista) (pack (tira lista)))
)
)
(defun pega (lista)
(cond ((eql lista nil) nil)
((eql (cdr lista) nil) lista)
((equal (car lista) (cadr lista))
(cons (car lista) (pega (cdr lista))))
(t (list (car lista)))
)
)
(defun tira (lista)
(cond ((eql lista nil) nil)
((eql (cdr lista) nil) nil)
((equal (car lista) (cadr lista))
(tira (cdr lista)))
(t (cdr lista))
)
)
But the Prolog folks still have a neat solution:
pack([],[]).
pack([X|Xs],[Z|Zs]) :- transfer(X,Xs,Ys,Z), pack(Ys,Zs).
transfer(X,[],[],[X]).
transfer(X,[Y|Ys],[Y|Ys],[X]) :- X \= Y.
transfer(X,[X|Xs],Ys,[X|Zs]) :- transfer(X,Xs,Ys,Zs).
And Haskell still makes us all look like chumps:
group (x:xs) = let (first,rest) = span (==x) xs
in (x:first) : group rest
group [] = []
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