Question

鉴于代码：

data Error a = Fail|Ok a
               deriving (Eq, Ord, Show)

split :: Int -> [a] -> (Error ([a],[a]))
split 0 list = Ok ([], list)
split n list
    | n < 0 = Fail
    | n > length (list) = Fail
    | otherwise = Ok (take n list, drop n list)

interleave ::  [a] ->  [a] -> [a]
interleave list [] = list
interleave [] list = list
interleave (x:xs) (y:ys) = x : y : interleave xs ys


shuffle :: [Int] -> [a] -> Error [a]

我如何编写函数shuffle，它将获取Int列表，并根据这些int拆分另一个列表。 int列表的示例是intList = [20,23,24,13]，其中shuffle将在第20个元素之后拆分列表，交错，在第23个元素之后拆分，交错，等等。

Answer 1

好的，你想要的基本上如下：

给定列表xs和索引[a1, a2, a3, ..]，在xs处拆分a1，交错，将其拆分为a2并交错，依此类推。

现在我们有两个功能：

step :: Int -> [a] -> [a]
step index xs = ??? -- something with split and interleave

shuffle :: [a]
shuffle [] xs = xs
shuffle (i:indices) xs = let newList = step i xs
                         in ??? -- something with recursion

尝试自己编写其余的这些功能。

step可以很容易地表达为let (x1, x2) = split index xs in interleave x1 x2。基本上，shuffle的其余部分可以写为shuffle indices newList。

Answer 2

我明白了：

shuffle [] xs = Ok (xs)
shuffle (x:xs) list = case (split x list) of
                           Fail        -> Fail
                           Ok (l1, l2) -> shuffle xs (interleave l1 l2)

基于Int列表对列表进行混洗

2 个答案: