将列表拆分为奇数和偶数 (Haskell)

Question

我需要创建一个函数，它将接受一个整数列表并将它们分成 2 个列表，一个是奇数，另一个是偶数

split :: [a] -> ([a], [a])
split = undefined

上面是函数基线，下面是我目前的尝试

 split :: [a] -> ([a], [a])
 split [] = ([],[])
 split (x:xs) | x mod 2 == 0 = ([],[x:split xs])
         | x mod 2 /= 0 = ([x:split xs],[])
         | otherwise = ([],[])

Answer 1

显式递归：

split :: Integral a => [a] -> ([a], [a])
split []        = ([], [])
split (x:xs)
  | even x      = (x:ys, zs)
  | otherwise   = (ys, x:zs)
  where
    (ys, zs) = split xs

隐式递归：

splitf :: Integral a => [a] -> ([a], [a])
splitf xs = foldr (\x (ys, zs) -> if even x then (x:ys, zs) else (ys, x:zs)) 
                  ([], []) xs

您可以将其简化为无点样式：

splitf2 :: Integral a => [a] -> ([a], [a])
splitf2 = foldr (\x (ys, zs) -> if even x then (x:ys, zs) else (ys, x:zs)) 
                ([], [])

我认为代码是不言自明的。如果有什么需要我进一步解释的，请告诉我。

将列表拆分为奇数和偶数 (Haskell)

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