99 questions/Solutions/6
From HaskellWiki
< 99 questions  Solutions(Difference between revisions)
(Added solution using Control.Arrows fan out operator.) 

(7 intermediate revisions by 6 users not shown) 
Revision as of 20:49, 30 November 2012
(*) Find out whether a list is a palindrome. A palindrome can be read forward or backward; e.g. (x a m a x).
isPalindrome :: (Eq a) => [a] > Bool isPalindrome xs = xs == (reverse xs)
isPalindrome' [] = True isPalindrome' [_] = True isPalindrome' xs = (head xs) == (last xs) && (isPalindrome' $ init $ tail xs)
Here's one to show it done in a fold just for the fun of it. Do note that it is less efficient then the previous 2 though.
isPalindrome'' :: (Eq a) => [a] > Bool isPalindrome'' xs = foldl (\acc (a,b) > if a == b then acc else False) True input where input = zip xs (reverse xs)
Another one just for fun:
isPalindrome''' :: (Eq a) => [a] > Bool isPalindrome''' = Control.Monad.liftM2 (==) id reverse
Or even:
isPalindrome'''' :: (Eq a) => [a] > Bool isPalindrome'''' = (==) Control.Applicative.<*> reverse
Here's one that does half as many compares:
palindrome :: (Eq a) => [a] > Bool palindrome xs = p [] xs xs where p rev (x:xs) (_:_:ys) = p (x:rev) xs ys p rev (x:xs) [_] = rev == xs p rev xs [] = rev == xs
Here's one using foldr and zipWith.
palindrome :: (Eq a) => [a] > Bool palindrome xs = foldr (&&) True $ zipWith (==) xs (reverse xs) palindrome' xs = and $ zipWith (==) xs (reverse xs)  same, but easier
isPalindrome list = take half_len list == reverse (drop (half_len + (len `mod` 2)) list) where len = length list half_len = len `div` 2 isPalindrome' list = f_part == reverse s_part where len = length list half_len = len `div` 2 (f_part, s_part') = splitAt half_len list s_part = drop (len `mod` 2) s_part'
Using Control.Arrows (&&&) fan out operator.
With monomorphism restriction:
isPalindrome1 xs = (uncurry (==) . (id &&& reverse)) xs
Point free with no monomorphism restriction:
isPalindrome1 = (uncurry (==) . (id &&& reverse))