简单的haskell拆分列表

Question

我有以下函数,它接受一个列表并返回在给定元素n处拆分的两个子列表.但是,我只需要将它分成两半,奇数长度列表具有更大的第一个子列表

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置像length(xs)这样的东西,这样我就不会破坏递归？谢谢.

Answer 1

你可以使用take和drop来做到这一点:

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功能:

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

Answer 2

在一个真正的程序中你应该使用

splitlist :: [a] -> ([a], [a])
splitlist xs = splitAt ((length xs + 1) `div` 2) xs

(即类似于dreamcrash的回答.)

但是,如果出于学习目的,您正在寻找明确的递归解决方案,请研究:

splitlist :: [a] -> ([a], [a])
splitlist xs = f xs xs where
    f (y : ys) (_ : _ : zs) =
        let (as, bs) = f ys zs
        in (y : as, bs)
    f (y : ys) (_ : []) = (y : [], ys)
    f ys [] = ([], ys)