Thomas Davie wrote:
In this interprettation, here's what I think is an O(1)
implementation:
...
rbPop :: Colour -> RBStack a -> RBStack a
rbPop c Empty = error "Empty Stack, can't pop"
rbPop c (More c' v asCs nextNonC)
| c == c' = asCs
| otherwise = rbPop c nextNonC
...
Your pop does
Matthew Brecknell wrote:
Matthew Eastman said:
i.e. popping Blue in [Red, Red, Blue, Red, Blue] would give [Red,
Red,
Blue]
Hmm, did you mean [Red,Blue] or [Red,Red,Red,Blue]? Judging by your
implementation of remUseless, I'm guessing the latter.
Yes, I meant the latter. Popping Blue in
On 25 Sep 2008, at 06:11, Matthew Eastman wrote:
Hey guys,
This is probably more of a question about functional programming
than it is about Haskell, but hopefully you can help me out. I'm new
to thinking about things in a functional way so I'm not sure what
the best way to do some thing
On Thu, 2008-09-25 at 00:11 -0400, Matthew Eastman wrote:
> Hey guys,
>
> This is probably more of a question about functional programming than
> it is about Haskell, but hopefully you can help me out. I'm new to
> thinking about things in a functional way so I'm not sure what the
> best way
Matthew Eastman said:
> i.e. popping Blue in [Red, Red, Blue, Red, Blue] would give [Red, Red,
> Blue]
Hmm, did you mean [Red,Blue] or [Red,Red,Red,Blue]? Judging by your
implementation of remUseless, I'm guessing the latter.
Here is a more straightforward approach than apfelmus'. I store colou
Try writing
data RBStack = RBS [RBSItem] [RBSItem]
where the first list are all the same colour and the start of the
second list is a different colour. The rest should follow naturally
and you will get amortised O(1) push and pop (you occasionally have to
juggle the lists).
By the way, for this
Hey guys,
This is probably more of a question about functional programming than
it is about Haskell, but hopefully you can help me out. I'm new to
thinking about things in a functional way so I'm not sure what the
best way to do some things are.
I'm in a data structures course right now,
