[Haskell-cafe] working with lists of couples
zacara at gmail.com
Thu Nov 23 17:51:27 EST 2006
Yes the code you are suggesting is certainly linear and it takes without
doubt time n.
But I was looking for a solution using foldl that of course takes time n.
The problem of the following solution is that it gives a reversed result,
and of course i can't just reverse the result.
couples = snd . foldl f (0,)
f (s,) x = (s+x, [(x,0)])
f (s,xs) x = (s+x, (:) (x,s) xs)
2006/11/17, Valentin Gjorgjioski <valentin.gjorgjioski at ijs.si>:
> On 17.11.2006 21:04 Clare wrote:
> > I'm not sure it takes time n couse of the operator ++ and the lazy
> > stuffs in haskell.
> Ok, you can use
> buildCouples (x:xs) s = (x,s) : (buildCouples xs (x+s))
> instead of ++
> this algorithm is linear, I don't know why(?) you think it is not.
> Valentin Gjorgjioski
> Bachelor of Computer Science
> Department of Knowledge Technologies, Jozef Stefan Institute
> Jamova 39, SI-1000 Ljubljana, Slovenia
> Phone: +386 1 477 3343
> Fax: +386 1 477 3315
> Web: http://kt.ijs.si/ValentinGjorgjioski/
> Email: Valentin.Gjorgjioski at ijs.si
-------------- next part --------------
An HTML attachment was scrubbed...
More information about the Haskell-Cafe