合并Haskell中的两个列表

merge :: [a] -> [a] -> [a]
merge xs     []     = xs
merge []     ys     = ys
merge (x:xs) (y:ys) = x : y : merge xs ys

我想提出一个更懒惰的合并版本:

merge [] ys = ys
merge (x:xs) ys = x:merge ys xs

例如,您可以查看最近的一个关于 lazy generation of combinations .
接受答案的版本在第二个论点中是不必要的严格,这就是这里所改进的。

所以你为什么这么简单(concat.转置)“不够漂亮”?我想你尝试过这样的方法:

merge :: [[a]] -> [a]
merge = concat . transpose

merge2 :: [a] -> [a] -> [a]
merge2 l r = merge [l,r]

因此,您可以避免显式递归(与第一个答案相比),但它仍然比第二个答案简单。那么缺点是什么呢?

编辑:看看艾德卡的回答和评论!

另一种可能性:

merge xs ys = concatMap (\(x,y) -> [x,y]) (zip xs ys)

或者,如果你喜欢实用的:

merge xs ys = concat $ getZipList $ (\x y -> [x,y]) <$> ZipList xs <*> ZipList ys

当然,这是一个展开的案例:

interleave :: [a] -> [a] -> [a]
interleave = curry $ unfoldr g
  where
    g ([], [])   = Nothing
    g ([], (y:ys)) = Just (y, (ys, []))
    g (x:xs, ys) = Just (x, (ys, xs))

-- ++
pp [] [] = []
pp [] (h:t) = h:pp [] t
pp (h:t) [] = h:pp t []
pp (h:t) (a:b) = h : pp t (a:b)