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我似乎不知道如何编写这个函数。我正在尝试编写的是一个函数expand，它将列表lst作为形式'(a (2 b) (3 c))的参数，并被计算为'(a b b c c c)<</p>
这看起来像家庭作业，所以我不会给你一个直接的答案。相反，我会给你一些正确方向的指示。最有用的提示是，您应该将问题拆分为两个过程，一个用于处理"外部"列表，另一个用于生成内部子列表中编码的重复项。
请注意，这两个过程是相互递归的（例如，它们相互调用）。expand过程在列表中重复，而repeat过程在重复次数上重复。这是所提出的解决方案的一般结构，填空：
; input: lst - list to be processed
; output: list in the format requested
(define (expand lst)
(cond ((null? lst) ; if the list is null
'()) ; then return null
((not (pair? (car lst))) ; if the first element of the list is an atom
(cons <???> <???>)) ; cons the atom and advance the recursion
(else ; if the first element of the list is a list
<???>))) ; call `repeat` with the right params
; input: n - number of repetitions for the first element in the list
; lst - list, its first element is of the form (number atom)
; output: n repetitions of the atom in the first element of lst
(define (repeat n lst)
(if (zero? n) ; if the number of repetitions is zero
(expand (cdr lst)) ; continue with expand's recursion
(cons <???> ; else cons the atom in the first element and
<???>))) ; advance the recursion with one less repetition
由于这是三年前回答的，我不认为我在帮助家庭作业。只想指出，这两个函数真的need不是相互递归的。由于replicate是一个相当常见的函数，我建议：
(define (replicate what n)
(if (zero? n)
(list)
(cons what (replicate what (- n 1)))))
(define (my-expand xs)
(if (empty? xs)
(list)
(let ((x (first xs)))
(if (list? x)
(let ((the-number (first x))
(the-symbol (cadr x)))
(flatten (cons (replicate the-symbol the-number)
(my-expand (rest xs)))))
(cons x (my-expand (rest xs)))))))
当然，最好使用两个列表并在最后执行平展，如下所示：
(define (my-expand xs)
(define (inner-expander xs ys)
(if (empty? xs) (flatten (reverse ys))
(let ((x (first xs)))
(if (list? x)
(let ((the-number (first x))
(the-symbol (cadr x)))
(inner-expander (rest xs) (cons (replicate the-symbol the-number) ys)))
(inner-expander (rest xs) (cons x ys))))))
(inner-expander xs (list)))