Personal tools

99 questions/Solutions/6

From HaskellWiki

< 99 questions | Solutions(Difference between revisions)
Jump to: navigation, search
Current revision (20:49, 30 November 2012) (edit) (undo)
(Added solution using Control.Arrows fan out operator.)
 
(2 intermediate revisions not shown.)
Line 32: Line 32:
<haskell>
<haskell>
isPalindrome'''' :: (Eq a) => [a] -> Bool
isPalindrome'''' :: (Eq a) => [a] -> Bool
-
ispalindrome'''' = (==) Control.Applicative.<*> reverse
+
isPalindrome'''' = (==) Control.Applicative.<*> reverse
</haskell>
</haskell>
Line 50: Line 50:
palindrome :: (Eq a) => [a] -> Bool
palindrome :: (Eq a) => [a] -> Bool
palindrome xs = foldr (&&) True $ zipWith (==) xs (reverse xs)
palindrome xs = foldr (&&) True $ zipWith (==) xs (reverse xs)
 +
palindrome' xs = and $ zipWith (==) xs (reverse xs) -- same, but easier
 +
</haskell>
 +
 +
 +
<haskell>
 +
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'
 +
</haskell>
 +
 +
 +
Using Control.Arrows (&&&) fan out operator.
 +
 +
With monomorphism restriction:
 +
 +
<haskell>
 +
isPalindrome1 xs = (uncurry (==) . (id &&& reverse)) xs
 +
</haskell>
 +
 +
Point free with no monomorphism restriction:
 +
 +
<haskell>
 +
isPalindrome1 = (uncurry (==) . (id &&& reverse))
</haskell>
</haskell>

Current revision

(*) 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))
Retrieved from "http://www.haskell.org/haskellwiki/99_questions/Solutions/6"