我一直在尝试找出一种算法,它可以实现以下功能:
算法将收到这样的列表:
((start a b c) (d e f (start g h i) (j k l) (end)) (end) (m n o))
然后,它将包含元素开始的列表与所有列表连接,直到包含元素结束。然后返回的列表应该是这样的:
((start a b c (d e f (start g h i (j k l)))) (m n o))
该算法必须能够处理包含start的列表以及包含start 编辑: 我现在拥有的是: 但它不起作用。也许我应该列出或追加而不是考虑? 编辑2: 上面的程序不应该工作,因为我正在尝试从非列表中获取第一个元素。这就是我到目前为止所想到的: 这就是我得到的结果:(defun conc-lists (l)
(cond
((endp l) '())
((eq (first (first l)) 'start)
(cons (cons (first (first l)) (conc-lists (rest (first l)))))
(conc-lists (rest l)))
((eq (first (first l)) 'end) '())
(t (cons (first l) (conc-lists (rest l))))))
(defun conc-lists (l)
(cond
((endp l) '())
((eq (first (first l)) 'start)
(append (cons (first (first l)) (rest (first l)))
(conc-lists (rest l))))
((eq (first (first l)) 'end) '())
(t (cons (first l) (conc-lists (rest l))))))
(conc-lists ((START A B C) (D E F (START G H I) (J K L) (END)) (END) (M N O)))
1. Trace: (CONC-LISTS '((START A B C) (D E F (START G H I) (J K L) (END)) (END) (M N O)))
2. Trace: (CONC-LISTS '((D E F (START G H I) (J K L) (END)) (END) (M N O)))
3. Trace: (CONC-LISTS '((END) (M N O)))
3. Trace: CONC-LISTS ==> NIL
2. Trace: CONC-LISTS ==> ((D E F (START G H I) (J K L) (END)))
1. Trace: CONC-LISTS ==> (START A B C (D E F (START G H I) (J K L) (END)))
(START A B C (D E F (START G H I) (J K L) (END)))
我也是CL的相对初学者,但这似乎是一个有趣的挑战,所以我尝试了一下。经验丰富的口译人员,请对此代码发表评论@user1176517,如果你发现任何错误,请告诉我!
首先有几条评论:我想让它成为O(n),而不是O(n^2),所以我让递归函数返回这两个列表的头和尾(即最后的cons),这两个都是递归处理树的分支所产生的。这样,在conc-lists-start
中,我可以将一个列表的最后一个cons nconc
放到另一个列表中的第一个cons上,而nconc
不必沿着列表走下去。我使用了多个返回值来做这件事,不幸的是,这让代码有点膨胀。为了确保tail
是结果列表的最后一个cons,我需要在重复出现之前检查cdr
是否为null。
有两个递归函数来处理树:conc-lists
和conc-lists-first
。当conc-lists
看到(start)
时,递归处理继续到conc-lists-start
。同样,当conc-lists-start
看到(end)
时,递归处理继续到conc-lists
。
我相信它可能需要更多的评论。。。我稍后可能会补充更多内容。
这是工作代码:
;;; conc-lists
;;; runs recursively over a tree, looking for lists which begin with 'start
;;; such lists will be nconc'd with following lists a same level of nesting,
;;; up until the first list which begins with 'end
;;; lists which are nconc'd onto the (start) list are first recursively processed
;;; to look for more (start)s
;;; returns 2 values: head *and* tail of resulting list
;;; DESTRUCTIVELY MODIFIES ARGUMENT!
(defun conc-lists (lst)
(cond
((or (null lst) (atom lst)) (values lst lst))
((null (cdr lst)) (let ((head (conc-process-rest lst)))
(values head head)))
(t (conc-process-rest lst))))
;;; helper to factor out repeated code
(defun conc-process-rest (lst)
(if (is-start (car lst))
(conc-lists-start (cdar lst) (cdr lst))
(multiple-value-bind (head tail) (conc-lists (cdr lst))
(values (cons (conc-lists (car lst)) head) tail))))
;;; conc-lists-start
;;; we have already seen a (start), and are nconc'ing lists together
;;; takes *2* arguments so that 'start can easily be stripped from the
;;; arguments to the initial call to conc-lists-start
;;; recursive calls don't need to strip anything off, so the car and cdr
;;; are just passed directly
(defun conc-lists-start (first rest)
(multiple-value-bind (head tail) (conc-lists first)
(cond
((null rest) (let ((c (list head))) (values c c)))
((is-end (car rest))
(multiple-value-bind (head2 tail2) (conc-lists (cdr rest))
(values (cons head head2) tail2)))
(t (multiple-value-bind (head2 tail2) (conc-lists-start (car rest) (cdr rest))
(nconc tail (car head2))
(values (cons head (cdr head2)) tail2))))))
(defun is-start (first)
(and (listp first) (eq 'start (car first))))
(defun is-end (first)
(and (listp first) (eq 'end (car first))))