有人请帮我这个功能吗?
使用方案功能,其行为类似于交换的递归版本。
(reswap '((h i)(j k) l (m n o)))
应该回来
((k j) (i h) (n m o) l) ;
和
(reswap '((a b) c (d (e f)) g (h i)))
应该回来
(c (b a) g ((f e) d) (i h)))
试试这个:
(define (rswap lst)
;; Create a helper function to do the recursive work.
(define (helper in out)
;; If the input is not a list, simply return it.
;; There is nothing to be done to rswap it.
(if (not (list? in))
in
;; If in is an empty list, simply return the out.
(if (null? in)
out
;; If in is a list with only one item, append
;; the result of calling rswap on the item to
;; out and return it.
(if (null? (cdr in))
(append out (list (rswap (car in))))
;; This is where the recursion continues.
;; Take two items off in before the next call.
;; rswap the two items and add them to out.
(helper
(cddr in)
(append out (list (rswap (cadr in)) (rswap (car in)))))))))
(helper lst '()))
本文收集自互联网,转载请注明来源。
如有侵权,请联系[email protected] 删除。
我来说两句