99 questions/Solutions/4

From HaskellWiki
< 99 questions‎ | Solutions
Revision as of 03:06, 16 July 2012 by Yiannist (talk | contribs) (NTUA solution ;-))
Jump to navigation Jump to search
The printable version is no longer supported and may have rendering errors. Please update your browser bookmarks and please use the default browser print function instead.

(*) Find the number of elements of a list.

myLength           :: [a] -> Int
myLength []        =  0
myLength (_:xs)    =  1 + myLength xs

myLength' :: [a] -> Int
myLength' list = myLength_acc list 0 -- same, with accumulator
	where
		myLength_acc [] n = n
		myLength_acc (_:xs) n = myLength_acc xs (n + 1)
myLength'     =  foldl (\n _ -> n + 1) 0
myLength''    =  foldr (\_ n -> n + 1) 0
myLength'''   =  foldr (\_ -> (+1)) 0
myLength''''  =  foldr ((+) . (const 1)) 0
myLength''''' =  foldr (const (+1)) 0
myLength' xs = snd $ last $ zip xs [1..] -- Just for fun
myLength''   = snd . last . (flip zip [1..]) -- Because point-free is also fun
myLength'''  = fst . last . zip [1..] -- same, but easier
myLength = sum . map (\x -> 1)
myLength = foldl (const . (+1)) 0

This is length in Prelude.