Question

我正在尝试将列表拆分为两个，以便在输入[1,2,3,5,6]时输出为[1,2,3][5,6]，但我似乎无法弄明白。我能做的最好的是[1,3,6][2,5]。

Answer 1

我是初学者。所以，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]

<强>输出

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

修改

管理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]

<强>输出

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

修改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]

<强>输出

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

<强> EDIT2

正如Bogdon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]

<强>输出

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

如何在haskell中将列表拆分为两个？

1 个答案: