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(Just wanted to post a solution using a fold.)
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isPalindrome' [_] = True
 
isPalindrome' [_] = True
 
isPalindrome' xs = (head xs) == (last xs) && (isPalindrome' $ init $ tail xs)
 
isPalindrome' xs = (head xs) == (last xs) && (isPalindrome' $ init $ tail xs)
  +
</haskell>
  +
  +
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.
  +
  +
<haskell>
  +
isPalindrome'' :: (Eq a) => [a] -> Bool
  +
isPalindrome'' xs = foldl (\acc (a,b) -> if a == b then acc else False) True input --just to be different
  +
where
  +
input = zip xs (reverse xs)
 
</haskell>
 
</haskell>

Revision as of 04:52, 28 February 2011

(*) 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 --just to be different
	where
	input = zip xs (reverse xs)