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,[])
where
f (s,[]) x = (s+x, [(x,0)])
f (s,xs) x = (s+x, (:) (x,s) xs)
Clare
2006/11/17, Valentin Gjorgjioski <[EMAIL PROTECTED]>:
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
--
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: [EMAIL PROTECTED]
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