Posted on 2008-04-16 10:27
dennis 阅读(606)
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计算机科学与基础
sicp的习题3.22，也就是以消息传递的风格重新实现队列，我的解答如下：
(define (make-queue)
(let ((front-ptr '())
(rear-ptr '()))
(define (set-front-ptr! ptr) (set! front-ptr ptr))
(define (set-rear-ptr! ptr) (set! rear-ptr ptr))
(define (empty-queue?) (null? front-ptr))
(define (front-queue)
(if (empty-queue?)
(error "FRONT called with an empty queue")
(car front-ptr)))
(define (insert-queue! item)
(let ((new-pair (cons item '())))
(cond ((empty-queue?)
(set-front-ptr! new-pair)
(set-rear-ptr! new-pair))
(else
(set-cdr! rear-ptr new-pair)
(set-rear-ptr! new-pair)))))
(define (delete-queue!)
(cond ((empty-queue?)
(error "DELETE! called with an empty queue" queue))
(else
(set-front-ptr! (cdr front-ptr)))))
(define (dispatch m)
(cond ((eq? m 'front-queue) (front-queue))
((eq? m 'empty-queue?) (empty-queue?))
((eq? m 'insert-queue!) insert-queue!)
((eq? m 'delete-queue!) delete-queue!)
(else
(error "Unknow method" m))))
dispatch))
(define (front-queue z) (z 'front-queue))
(define (empty-queue? z) (z 'empty-queue?))
(define (insert-queue! z item) ((z 'insert-queue!) item))
(define (delete-queue! z) ((z 'delete-queue!)))
由此，我才知道自己竟然一直没有想到，scheme完全可以模拟单向循环链表，整整第三章都在讲引入赋值带来的影响，而我却视而不见。在引入了改变函数后，数据对象已经具有OO的性质，模拟链表、队列、table都变的易如反掌。首先，模拟节点对象，节点是一个序对，包括当前节点编号和下一个节点：
(define (make-node n next) (cons n next))
(define (set-next-node! node next) (set-cdr! node next))
(define (set-node-number! node n) (set-car! node n))
(define (get-number node) (car node))
(define (get-next-node node) (cdr node))
有了节点，实现了下单向循环链表：
(define (make-cycle-list n)
(let ((head (make-node 1 '())))
(define (make-list current i)
(let ((next-node (make-node (+ i 1) '())))
(cond ((= i n) current)
(else
(set-next-node! current next-node)
(make-list next-node (+ i 1))))))
(set-next-node! (make-list head 1) head)
head))
make-cycle-list生成一个有N个元素的环形链表，比如(make-cycle-list 8)的结果如下
#0=(1 2 3 4 5 6 7 8 . #0#)
Drscheme形象地展示了这是一个循环的链表。那么约瑟夫环的问题就简单了：
(define (josephus-cycle n m)
(let ((head (make-cycle-list n)))
(define (josephus-iter prev current i)
(let ((next-node (get-next-node current)))
(cond ((eq? next-node current) (get-number current))
((= 1 i)
(set-next-node! prev next-node)
(josephus-iter prev next-node m))
(else
(josephus-iter current next-node (- i 1))))))
(josephus-iter head head m)))
从head节点开始计数，每到m，就将当前节点删除（通过将前一个节点的next-node设置为current的下一个节点），最后剩下的节点的编号就是答案。