简单的haskell splitlist

问题描述

我有下面的函数，它需要一个列表并返回在给定元素n处分割的两个子列表。然而，我只需要将它分成两半，奇数长度的列表具有较大的第一个子列表。

I have the following function which takes a list and returns two sublists split at a given element n. However, I only need to split it in half, with odd length lists having a larger first sublist

splitlist :: [a] -> Int -> ([a],[a])
splitlist [] = ([],[])
splitlist l@(x : xs) n | n > 0     = (x : ys, zs)
               | otherwise = (l, [])
    where (ys,zs) = splitlist xs (n - 1)

我知道我需要将签名更改为[a] - >（[a]，[a]），但是在代码中我应该放入类似length（xs）的东西，以便不会中断递归？谢谢。

I know I need to change the signature to [a] -> ([a],[a]), but where in the code should I put something like length(xs) so that I don't break recursion? Thank you.

推荐答案

您可以使用take and drop：

You can do it using take and drop:

splitlist :: [a] -> ([a],[a])
splitlist [] = ([],[])
splitlist l  = let half = (length(l) +1)`div` 2
               in (take half l, drop half l)

或者您可以利用splitAt函数：

or you can take advantage of the function splitAt:

splitlist list = splitAt ((length (list) + 1) `div` 2) list

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