Get all permutations of a list in Haskell

Question

I\'m trying to do this from scratch, without the use of a library outside the standard lib. Heres my code:

permutations :: [a] -> [[a]]
permutations (x:xs         


        

Answer 1

Everything is better with monads:

perm :: [a] -> [[a]]
perm []     = return []
perm (x:xs) = (perm xs) >>= (ins x)
    where
    ins :: a -> [a] -> [[a]]
    ins x []     = [[x]]
    ins x (y:ys) = [x:y:ys] ++ ( map (y:) (ins x ys) )

So: you have a function, that inserts letter in a word, but it produces more then one word, so how to apply it recursively? >>= helps!

Answer 2

I think that this is shorter and more elegant variant for what others are suggesting:

permutate :: (Eq a) => [a] -> [[a]]
permutate [] = [[]]
permutate l = [a:x | a <- l, x <- (permutate $ filter (\x -> x /= a) l)]