Question

我是Haskell的新手，我找到了一些练习的例子，但不知道如何处理它们。我需要将"abcd"分成子序列。

为：

["",a,b,c,d,ab,ac,ad,bc,bd,cd,abc,abd,bcd,acd,abcd]

我发现我必须使用子序列函数。

subsequences :: [a] -> [[a]]

我该怎么办？

Answer 1

您可以按照建议subsequences进行操作，看一下示例：

import Data.List

subsequences "abcd"

<强>输出：

["","a","b","ab","c","ac","bc","abc","d","ad","bd","abd","cd","acd","bcd","abcd"]

注意： subsequences不对输出进行排序。但是，如果您想要排序的输出，可以使用以下内容：

import Data.List
import Data.Function

sortBy (compare `on` length) $ subsequences "abcd"

或

import Data.List import GHC.Exts sortWith length $ subsequences "abcd"

<强>输出：

["","a","b","c","d","ab","ac","bc","ad","bd","cd","abc","abd","acd","bcd","abcd"]

点击此处了解更多信息：

Data.List.subsequences

Answer 2

这是一个简单的递归任务。

假设您有一个子列表的子序列列表。如何生成较长列表的子序列列表？

我建议将子序列列表按其长度分成子序列列表。所以你有[[subsequences length 0],[subsequences length 1],...]。很容易看出，通过在较短的子序列中附加一个元素可以产生更长的子序列：[[subsequences length 0],[subsequences length 0 plus one element]++[subsequences length 1],[subsequences length 1 plus one element]++[subsequences length 2],...]。

要对齐不同长度的子序列列表，请在子序列列表中添加一个空列表。然后将一个元素附加到较短的子序列，并与较长的子序列列表连接。

addoneall x xs = zipWith ((++) . (map (x:))) ([]:xs) (xs ++ [[]])

现在

subsequences = concat . foldr addoneall [[[]]]

生成所有长度的子序列列表 - foldr通过将一个元素附加到较短输入的子序列列表来生成子序列列表。然后concat将不同长度的子序列列表连接成一个。

Prelude Data.List> subsequences "abcd"
["","a","b","c","d","ab","ac","ad","bc","bd","cd","abc","abd","acd","bcd","abcd"]

Haskell：String的子序列

2 个答案: