问题
Someone please help me with this function?
use scheme function that behaves like a recursive version of swap.
(reswap '((h i)（j k) l (m n o)))
should return
((k j) (i h) (n m o) l) ;
and
(reswap '((a b) c (d (e f)) g (h i)))
should return
(c (b a) g ((f e) d) (i h)))
回答1:
Try this:
(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 '()))
回答2:
Lol this looks like a good question but all I got was
(define (swap lst)
; if the list is empty or has a single element
(cond ((or (null? lst) (null? (cdr lst)))
; then return that list
lst)
(else
; by first adding the second element
(cons (cadr lst)
(cons (car lst)
(swap (cddr lst)))))))
but that just does a normal swap.
来源：https://stackoverflow.com/questions/22902820/scheme-rswap-function