Haskell中的所有列表循环 [英] All list rotations in Haskell

问题描述

我有一个函数来旋转列表：

  rotate :: [a]  - > [a] 
 rotate [] = [] 
 rotate（x：xs）= xs ++ [x] 
  

现在我想要一个函数给出一个列表，列出每个可能的有限列表旋转：

  rotateAll :: [a]  - >在命令式语言中，我会做类似于（以伪代码的形式）的内容。[/ b] [b] [b] [b] [b]对于i = 1到列表的长度
追加列表到rotateList 
 list = rotate（list）
 
 $ b 
  $ b  
当然，必要的思考可能不会帮助我找到解决此问题的功能性解决方案。我正在寻找一些关于如何解决这个问题的提示。
 
 
 
 其他想法：
 
 
为了解决这个问题，我有两个问题需要解决。首先，我需要重复旋转一个列表并将每个结果收集到一个列表中。因此，第一个解决方案需要做一些类似于 
 
 
  rotateAll xs = [xs（rotate xs）（rotate（rotate xs））（旋转（旋转（旋转xs）））...] 
  
当然，我不知道如何很多时候这样做。我会满意地做到这一点，然后使用 take（length xs）来获得我想要的有限数量的列表。这实际上证明了第二个问题：确定何时停止。我不知道使用取是解决问题的最有效或最优雅的方法，但是当我输入这个内容时它就浮现出来了，应该可以工作。
 
 
 
 附录： 
现在我已经找到了两种解决方案，我很乐意欢迎任何其他更快或使用不同方法的解决方案。谢谢！
解决方案
除迭代，你可以编写一个函数来生成n个循环的列表。情况n = 0会将输入包装在一个列表中，并且情况n = m + 1会将输入附加到情况m的结果中。虽然使用标准功能通常是首选，但有时编写自己的解决方案是健康的。 
I have a function to rotate a list:
rotate :: [a] -> [a]
rotate [] = []
rotate (x:xs) = xs ++ [x]
Now I want a function that gives a list with every possible rotation of a finite list:
rotateAll :: [a] -> [[a]]
In a imperative language, I would do something like (in pseudocode)
for i = 1 to length of list
  append list to rotateList
  list = rotate(list)
Of course, thinking imperatively probably doesn't help me find a functional solution to this problem. I'm looking for some hints as to how to tackle this.

Additional thoughts:

To solve this, I have two issues to tackle. First I need to repeatedly rotate a list and collect each result into a list. So a first solution needs to do something like
rotateAll xs = [xs (rotate xs) (rotate (rotate xs)) (rotate (rotate (rotate xs))) ...]
Of course I don't know how many times to do this. I'd be satisfied to do this infinitely and then use take (length xs) to get the finite number of lists I desire. This actually demonstrates the second issue: determining when to stop. I don't know if using take is the most efficient or elegant way to solve the problem, but it came to mind as I was typing this and should work.

Addendum:
Now that I have found two solutions on my own or with hints. I will gladly welcome any other solutions that are faster or use different approaches. Thanks!
解决方案
Aside from iterate, you could write a function that generates a list of n rotations. Case n=0 would just wrap the input in a list and case n=m+1 would append the input to the result of case m. Although using standard functions is generally preferred, sometimes writing your own solutions is healthy.

                        
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