方案-生成列表的所有不同排列

Question

[特别是读一本有关函数式编程和方案（以及Racket）的书时，我碰巧遇到了一个练习，陈述如下：`

"Write a function 'rp' which takes, as an argument, a list 'lp' of pairs '(a . n)',
where 'a' is either a symbol or a number and 'n' is a natural number, 
and which returns the list of all the lists, whose elements are the 'a's defined by 
the pairs in 'lp', each one appearing exactly 'n' times."

出于某种原因，这确实是个谜，但它的基本要求是包含n乘以数字/符号a的列表的所有不同排列的列表。

例如：[[(rp '((a . 2) (b . 1))]] = '((a a b) (a b a) (b a a))

[生成排列而忽略distinct部分，是相当容易的，因为有一个相对简单的递归定义：

The list of permutations of an empty list, is a list containing an empty list.
The list of permutations of 3 elements a b c is a list containing the lists of all permutations of
a and b where, for each one, c has been inserted in all possible positions.

我翻译了以下球拍代码：

(define permut
  (lambda(ls)
    (if(null? ls) '(())
       (apply append
              (map (lambda(l) (insert_perm (car ls) l))
                   (permut (cdr ls)))))))

(define insert_perm
  (lambda(x ls)
    (if(null? ls) (list (list x))
       (cons (cons x ls)
             (map (lambda(l) (cons (car ls) l))
                  (insert_perm x (cdr ls)))))))

这有效，但不会返回不同的排列。在我看来，考虑重复项似乎要复杂得多。我看不到简单排列情况的简单修改吗？解决方案完全不同吗？任何帮助，将不胜感激。

Answer 1

; When your input is '((a 2) (b 1))
; Or something like that. it can be (a . 2) or vector('a 2)
; change input become normal list -> L = '(a a b) 

; just use permutations -> (permutations input-L)
; If you don't want repeate just remove repeate. -> (remove-duplicates output-List)
; Done.

更新2