99 questions/Solutions/4
From HaskellWiki
(Difference between revisions)
(Added another two solutions) |
|||
| Line 5: | Line 5: | ||
myLength [] = 0 | myLength [] = 0 | ||
myLength (_:xs) = 1 + myLength xs | 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) | ||
</haskell> | </haskell> | ||
| Line 18: | Line 24: | ||
myLength' xs = snd $ last $ zip xs [1..] -- Just for fun | myLength' xs = snd $ last $ zip xs [1..] -- Just for fun | ||
myLength'' = snd . last . (flip zip [1..]) -- Because point-free is also fun | myLength'' = snd . last . (flip zip [1..]) -- Because point-free is also fun | ||
| + | myLength''' = fst . last . zip [1..] -- same, but easier | ||
</haskell> | </haskell> | ||
Revision as of 04:45, 24 June 2012
(*) 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)
length
Prelude
