Difference between revisions of "99 questions/Solutions/19"
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or calculate the
or calculate the (index) at first:
Revision as of 07:07, 12 August 2017
(**) Rotate a list N places to the left.
Hint: Use the predefined functions length and (++).
rotate  _ =  rotate xs 0 = xs rotate (x:xs) (n+1) = rotate (xs ++ [x]) n rotate xs n = rotate xs (length xs + n)
There are two separate cases:
- If n > 0, move the first element to the end of the list n times.
- If n < 0, convert the problem to the equivalent problem for n > 0 by adding the list's length to n.
or using cycle:
rotate xs n = take len . drop (n `mod` len) . cycle $ xs where len = length xs
or using list comprehension (only works for sequential increasing elements):
rotate :: (Enum a) => [a] -> Int -> [a] rotate xs n = [(f n) .. last xs] ++ [head xs .. (f (n-1))] where f k = xs !! (k `mod` length xs)
or without mod:
rotate xs n = take (length xs) $ drop (length xs + n) $ cycle xs
rotate xs n = if n >= 0 then drop n xs ++ take n xs else let l = ((length xs) + n) in drop l xs ++ take l xs
rotate xs n | n >= 0 = drop n xs ++ take n xs | n < 0 = drop len xs ++ take len xs where len = n+length xs
or calculate the position(index) at first:
rotate xs n = let i = if n < 0 then length xs + n else n in drop i xs ++ take i xs
rotate xs n = drop nn xs ++ take nn xs where nn = n `mod` length xs
Using a simple splitAt trick
rotate xs n | n < 0 = rotate xs (n+len) | n > len = rotate xs (n-len) | otherwise = let (f,s) = splitAt n xs in s ++ f where len = length xs
rotate xs n | n > 0 = (reverse . take n . reverse $ xs) ++ (reverse . drop n . reverse $ xs) | n <= 0 = (drop (negate n) xs) ++ (take (negate n) xs)
A much simpler solution without using
length that is very similar to the first solution:
rotate :: [a] -> Int -> [a] rotate  _ =  rotate x 0 = x rotate x y | y > 0 = rotate (tail x ++ [head x]) (y-1) | otherwise = rotate (last x : init x) (y+1)