String Haskell：字符串的子序列

我是Haskell的新手，我找到了一些可以练习的例子，但不知道如何处理它们。 我需要将

“abcd”

拆分为子序列

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

致：

我发现我必须使用子序列函数

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

---

我应该如何继续？

您可以使用建议的

子序列来执行此操作。请查看示例：

import Data.List

subsequences "abcd"

输出：

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

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

注意：
子序列
不会对输出进行排序。但是，如果需要排序的输出，可以使用以下选项：

import Data.List
import Data.Function

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

或

import Data.List 
import GHC.Exts

sortWith length $ subsequences "abcd"

输出：

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

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

查看此处了解更多信息：



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为子序列列表。所以您有
[[subsequencelength 0]，[subsequencelength 1]，…]
。很容易看出，通过将一个元素添加到较短的子序列中，可以生成较长的子序列：
[[子序列长度0]，[子序列长度0加一个元素]++[子序列长度1]，[子序列长度1加一个元素]++[子序列长度2]，…]

要对齐不同长度的子序列列表，请将空列表添加到子序列列表中。然后将一个元素附加到较短的子序列，并使用较长的子序列列表连接

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

现在

生成所有长度的子序列列表-
foldr
通过在子序列列表中添加一个元素以获得较短的输入，生成子序列列表。然后，
concat
将不同长度的子序列列表合并为一个

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

是的，我知道。。但是这使得结果与上面所示的不一样。。我想我必须做一些其他的事情，但在我读到的章节中没有显示。只需要获取字符串并将其子序列写入特定的格式。。我是begginer:）一个简单的方法可以是（近似！）
排序（比较长度）$subsequences“abcd”
，但请自己尝试一种直接的方法…您确定序列以
…，bcd，acd，abcd]
结束，而不是
…，acd，bcd，abcd]
？是的，这是我所做的，但我需要得到这个输出：[“，a，b，c，d，ab，ac，ad，bc，bd，cd，abc，abd，bcd，acd，abc]@user2497254更新了答案以包含排序方法。让我们调试它，好吗？：）首先打开
ghci
并尝试逐行复制粘贴代码，如果有问题，请告诉我。@user2497254在.hs文件中不能有“顶级裸表达式”。将其转换为一个定义，如
sortedSubsequences=sortBy（compare`on`length）。子序列
，将文件加载到ghci中，并使用排序的子序列“abcd”进行测试。子序列