如何制作子串列表？ [英] How do I make a list of substrings?

问题描述

我试图列出所有子字符串，其中每个子字符串只有一个少于原始字符串的元素。

例如1234会导致[1234，123，12，1]

我希望只使用前奏（无导入）来实现这一点，所以不能使用子序列。

我是Haskell的新手，我知道一些问题与我的代码，但目前不知道如何解决它们。

  slist :: String  - > [String] 
 slist（x：xs）=（take（length（x：xs））（x：xs））++ slist xs 
  

如何递归地使用

编辑：想要通过递归地使用init

解决方案

  slist :: String  - > [String] 
 slist [] = [] 
  -  slist xs = [xs] ++（slist $ init xs）
 slist xs = xs：（slist $ init xs）
 
 main = do 
 print $ slist1234
  

I am trying to make a list of all substrings where each substring has one less element of the originial string.

e.g "1234" would result in ["1234","123","12","1"]

I would like to achieve this only using prelude (no import) so cant use subsequences.

I am new to Haskell, and I know some of the problems with my code but don't currently know how to fix them.

slist :: String -> [String]
slist (x:xs) = (take (length (x:xs)) (x:xs)) ++ slist xs

How can I do this recursively using

Edit: would like to this by using init recursively

解决方案

slist :: String -> [String]
slist [] = []
-- slist xs = [xs] ++ (slist $ init xs)
slist xs = xs : (slist $ init xs)

main = do 
    print $ slist "1234"

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