Question

我一直在研究常见的lisp，我已经阅读了示例和一些pdf，我的教授为作业分配了几个问题，我只能使用cons，append，list，equal，defun，car，cdr，cond之一。是要删除特定数量的列表，例如(remove 2 '(1 2 3 4)，结果：(1 3 4)，我的代码：

                                 (define (delete ele listx )
                                    (cond ( null? listx)
                                         '( ))
                                       ( ( equal? ele (car listx))
                                       (delete ele (cdr listx)))
                                  (else 
                                    (cons (car listx)
                                       (delete ele (cdr listx))))))

这并不难，但是现在我必须创建一个删除两个元素之间的函数，例如-> (delete'(a b c d e f) 2 5)；在第三和第四（a b e f）之间删除

我还没有尝试过任何东西，只是进行了研究，我开始编写一些代码，但是不确定

( defun deletemiddle (index1 index2 listx)
   (cond (( null listx) 0 )
        (( = valor1 (1) )
          ((cons (car listx)

Answer 1

如果起始索引为0或更小且结束索引大于0，则删除元素并递归
否则，限制元素并递归

递归时，列表的cdr和两个索引都递减1。

Answer 2

这里有其他痕迹，可以帮助找出如何递归构建结果列表：

  0: (DEL (1 2 3 4 5 6) 2 5)
    1: (DEL (2 3 4 5 6) 1 4)
      2: (DEL (3 4 5 6) 0 3)
        3: (DEL (4 5 6) 0 2)
          4: (DEL (5 6) 0 1)
          4: DEL returned (5 6)
        3: DEL returned (5 6)
      2: DEL returned (5 6)
    1: DEL returned (2 5 6)
  0: DEL returned (1 2 5 6)

  0: (DEL (1 2 3 4 5 6) 2 8)
    1: (DEL (2 3 4 5 6) 1 7)
      2: (DEL (3 4 5 6) 0 6)
        3: (DEL (4 5 6) 0 5)
          4: (DEL (5 6) 0 4)
            5: (DEL (6) 0 3)
              6: (DEL NIL 0 2)
              6: DEL returned NIL
            5: DEL returned NIL
          4: DEL returned NIL
        3: DEL returned NIL
      2: DEL returned NIL
    1: DEL returned (2)
  0: DEL returned (1 2)

  0: (DEL (1 2 3 4 5 6) 0 3)
    1: (DEL (2 3 4 5 6) 0 2)
      2: (DEL (3 4 5 6) 0 1)
      2: DEL returned (3 4 5 6)
    1: DEL returned (3 4 5 6)
  0: DEL returned (3 4 5 6)

Answer 3

(defun deletemiddle (list idx1 idx2 &optional (curr 0))
  (cond ((null list) '())
        ((< idx1 curr idx2) 
         (deletemiddle (cdr list) idx1 idx2 (1+ curr)))
        (t (cons (car list) 
                 (deletemiddle (cdr list) idx1 idx2 (1+ curr))))))

;; (< idx1 curr idx2) is short for: (and (> curr idx1) (< curr idx2))

(deletemiddle '(a b c d e f) 2 5)
;; => (A B C F)

尾部递归

(defun deletemiddle (list idx1 idx2 &optional (acc '()))
  (cond ((null list) (nreverse acc))
        ((and (<= idx1 -1) (> idx2 0))
         (deletemiddle (cdr list) (1- idx1) (1- idx2) acc))
        (t 
         (deletemiddle (cdr list) (1- idx1) (1- idx2) (cons (car list) acc))))))

(deletemiddle '(a b c d e f) 2 5)
;; => (A B C F)

常见的Lisp锻炼

3 个答案: