本文介绍了方案 Rswap 功能的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

有人帮我做这个功能吗?

Someone please help me with this function?

使用表现类似于交换递归版本的方案函数.

use scheme function that behaves like a recursive version of swap.

(reswap '((h i)(j k) l (m n o)))

应该返回

((k j) (i h) (n m o) l) ;

(reswap '((a b) c (d (e f)) g (h i)))

应该返回

(c (b a) g ((f e) d) (i h))) 

推荐答案

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 '()))

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10-14 08:08