Haskell:将偶数和奇数元素拆分为元组

我不能使用高阶函数。我只是无法弄清楚如何做到这一点。我对haskell很陌生。它也必须是递归的。

split :: [Int] -> ([Int],[Int])
split xs =

我被赋予了这个开始。老实说，我什至不知道从哪里开始解决这个问题。

例子:

split []
([],[])

split [1]
([1],[])

split [1,2,3,4,5,6,7,8,9,10]
([1,3,5,7,9],[2,4,6,8,10])

任何帮助将非常感激。

编辑:它的偶数和奇数位置。

所以

拆分 [3,6,8,9,10] 将是
([3,8,10],[6,9])

好的，所以我想出了这个。它不漂亮，但它似乎工作正常。

split :: [Int] -> ([Int],[Int])
split [] = ([],[])
split [xs] = ([xs],[])
split xs = (oddlist xs, evenlist xs)

oddlist :: [Int] -> ([Int])
oddlist xs | length xs <= 2 = [head(xs)]
           | otherwise = [head(xs)] ++ oddlist(tail(tail(xs)))

evenlist :: [Int] -> ([Int])
evenlist xs | length xs <= 3 = [head(tail(xs))]
            | otherwise = [head(tail(xs))] ++ evenlist(tail(tail(xs)))

最佳答案

split [] = ([], [])
split [x] = ([x], [])
split (x:y:xs) = (x:xp, y:yp) where (xp, yp) = split xs

关于Haskell:将偶数和奇数元素拆分为元组，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/3707753/