我试图写一个Common Lisp的功能，这将使我一个列表的所有可能的排列，使用每个元素只有一次。例如，列表'（1 2 3）将得到的输出（（1 2 3）（1 3 2）（2 1 3）（2 3 1）（3 1 2）（3 2 1））。

I'm trying to write a Common Lisp function that will give me all possible permutations of a list, using each element only once. For example, the list '(1 2 3) will give the output ((1 2 3) (1 3 2) (2 1 3) (2 3 1) (3 1 2) (3 2 1)).

我已经写的东西，那种作品，但它的笨重，它并不总是工作，我甚至不真正了解它。我不要求为code，说不定关于如何去想一些指导。我不知道很多关于编写算法。

I already wrote something that kind of works, but it's clunky, it doesn't always work and I don't even really understand it. I'm not asking for code, just maybe for some guidance on how to think about it. I don't know much about writing algorithms.

谢谢， 杰森

推荐答案

作为一个基本方法，所有排列遵循这个递归模式：

As a basic approach, "all permutations" follow this recursive pattern:

  all permutations of a list L is:
    for each element E in L:
      that element prepended to all permutations of [ L with E removed ]

如果我们把所给，你必须在你的列表中没有重复的元素，下面应该做的：

If we take as given that you have no duplicate elements in your list, the following should do:

(defun all-permutations (list)
  (cond ((null list) nil)
        ((null (cdr list)) (list list))
        (t (loop for element in list
             append (mapcar (lambda (l) (cons element l))
                            (all-permutations (remove element list)))))))