如何在保留字符串形状的同时旋转字符串列表中的字母？

Question

我正在尝试编写一个函数join :: [[a]] -> [[a]]，它将列表中列表的最后一个字符与列表中列表的第一个字符连接起来。

join ["m"] = ["m"]
join ["a","b","c"] = ["b","c","a"]
join ["goodday", "world", "i", "love", "haskell"] = ["ooddayw","orldi","l","oveh","askellg"]
join ["the", "catcher", "in", "the", "rye"] = ["hec","atcheri","nt","her","yet"]

我正在尝试编写一个仅使用 Haskell 中的基本函数（没有库函数）并且仅使用递归的代码。

但是，我似乎无法实现一段正常工作的代码。这是我到目前为止的代码：

join :: [[a]] -> [[a]]
join [[a]] = [[a]]
join (n:ns:nss) | null nss == False = ((i n ns) : k (ns:nss))
                | otherwise = []

可以这样做吗？

Answer 1

这是一个具有高阶函数的解决方案，在整个数据-操作范式中工作：

import Control.Applicative (liftA2)
import Data.List (unfoldr)
import Control.Arrow ( (>>>) )

rotateds :: [[a]] -> [[a]]
rotateds =
  map (splitAt 1)                     -- 1., 2.,
  >>> concatMap (\(a,b) -> [a,b])     -- 3.,
  >>> liftA2 (++) (drop 1) (take 1)   -- 4.,
  >>> unfoldr (Just . splitAt 2)      -- 5., 
  >>> takeWhile (not . null)          -- 5.,
  >>> map (\[a,b] -> (++) a b)        -- 6.

它通过了你所有的测试。所以是的，这是可能的。它的工作原理是：

1. turn each sublist in the input  [    [a,b,c,...]  ,  [d,......]  , ... ]
2. into a pair                     [ ([a],[b,c,...]) , ([d], [...]) , ... ]
3. splice'em in                    [  [a],[b,c,...]  ,  [d], [...]  , ... ]
4. move the first letter over      [      [b,c,...]  ,  [d], ........ [a] ]
5. and restore the structure back to how it was 
    by reconstituting the pairs    [     [[b,c,...]  ,  [d]], ........... ]
6. and appending them together     [     [ b,c,...   ,   d ], ........... ]

将其转换为直接手动递归是一项艰巨的任务留给狂热的学习者作为练习。