Swap two elements in a list by its indices

Question

Is there any way to swap two elements in a list if the only thing I know about the elements is the position at which they occur in the list.

To be more specific, I a

Answer 1

I really like @dfeuer 's solution. However there's still room for optimization by way of deforestation:

swap' :: Int -> Int -> [a] -> [a]
swap' first second lst = beginning $ [y] ++ (middle $ [x] ++ end)
  where
    (beginning, (x : r)) = swapHelp first lst
    (middle, (y : end)) = swapHelp (second - first - 1) r

swapHelp :: Int -> [a] -> ([a] -> [a],[a])
swapHelp 0 l     = (    id , l)
swapHelp n (h:t) = ((h:).f , r) where
                   (     f , r) = swapHelp (n-1) t