haskell - 如何在 Haskell 中将列表一分为二？

Question

我正在尝试将列表分成两部分，以便当输入为

[1,2,3,5,6]

输出将是

[1,2,3][5,6] 

但我似乎无法弄清楚。

我能做的最好的就是[1,3,6][2,5]。

Answer 1

我是初学者。因此，如果这是错误的或次优的，请纠正我。

internalSplit :: [a] -> Int -> [a] -> [[a]]
split :: [a] -> [[a]]

internalSplit (first:rest) count firstPart
    | count == 0 = [firstPart, (first:rest)]
    | otherwise  = internalSplit rest (count - 1) (firstPart ++ [first])

split myList =
    let listLength = length myList
    in
        if listLength `mod` 2 == 0 then
            internalSplit myList (listLength `div` 2) []
        else
            internalSplit myList ((listLength `div` 2) + 1) []

main = do
        print $ split [1, 2, 3, 5, 6]
        print $ split [1, 2, 3, 4, 5, 6]

输出

[[1,2,3],[5,6]]
[[1,2,3],[4,5,6]]

编辑：

设法使用内置函数并想出了这个

internalSplit :: [a] -> Int -> [[a]]
split :: [a] -> [[a]]

internalSplit myList splitLength = [(take splitLength myList), (drop splitLength myList)]

split myList =
    let listLength = length myList
    in
        if listLength `mod` 2 == 0 then
            internalSplit myList (listLength `div` 2)
        else
            internalSplit myList ((listLength `div` 2) + 1)

main = do
        print $ split [1, 2, 3, 5, 6]
        print $ split [1, 2, 3, 4, 5, 6]

输出

[[1,2,3],[5,6]]
[[1,2,3],[4,5,6]]

编辑1：

internalSplit :: [a] -> Int -> ([a], [a])
split :: [a] -> ([a], [a])

internalSplit myList splitLength = splitAt splitLength myList

split myList =
    let listLength = length myList
    in
        if listLength `mod` 2 == 0 then
            internalSplit myList (listLength `div` 2)
        else
            internalSplit myList ((listLength `div` 2) + 1)

main = do
        print $ split [1, 2, 3, 5, 6]
        print $ split [1, 2, 3, 4, 5, 6]

输出

([1,2,3],[5,6])
([1,2,3],[4,5,6])

编辑2

正如博格登在评论部分所建议的那样，这可以大大简化为

split :: [a] -> ([a], [a])
split myList = splitAt (((length myList) + 1) `div` 2) myList
main = do
        print $ split [1, 2, 3, 5, 6]
        print $ split [1, 2, 3, 4, 5, 6]

输出

([1,2,3],[5,6])
([1,2,3],[4,5,6])