Question

我遇到多级递归问题。我有两个号码n1和n2。当counter介于n1和n2之间时，我希望结果中包含数字，保持列表内部结构。

(defun izdvoj (lista n1 n2 counter)
  (cond ((null lista) counter)
        ((and (atom (car lista))
              (< counter n1)
              (izdvoj (cdr lista) n1 n2 (+ counter 1))))
        ((and (atom (car lista))
              (> counter n2)
              (izdvoj (cdr lista) n1 n2 (+ counter 1))))
        ((atom (car lista))
         (cons (car lista) (izdvoj (cdr lista) n1 n2 (+ counter 1))))
        (t
         (cons (izdvoj (car lista) n1 n2 counter) (izdvoj (cdr lista) n1 n2 counter)))))

(izdvoj '(1 2 (3 (4) 5 (6 (7)))(8 (9 (10 ((11))) 12)) (13 ((14) (15)))) 7 13 0)

结果应为((((7))) (8 (9 (10 ((11))) 12)) (13))

我得到(((4) ((7))) (((((11))) 12)) (((14) (15))))

有什么建议吗？

Answer 1

最简单的方法是返回索引以及帮助程序中的结果。这样，您知道如何在完成car并准备好cdr之后继续：

(defun get-range (list from-index to-index)
  (labels ((aux (list cur acc)
             (cond ((null list)
                    (values (nreverse acc) cur))
                   ((not (atom (car list)))
                    (multiple-value-bind (res cur)
                                         (aux (car list) cur '())
                      (aux (cdr list) cur (cons res acc))))
                   ((<= from-index cur to-index)
                    (aux (cdr list) (1+ cur) (cons (car list) acc)))
                   (t (aux (cdr list) (1+ cur) acc)))))
    (nth-value 0 (aux list 1 '()))))

(get-range '(a b (c d (e (f))) (g h)) 3 5)
; ==> ((c d (e ())) ())

lisp中的多级递归

1 个答案: