问题
Write a recursive Scheme procedure count-dist elements that takes a list with duplicate elements and returns the number of distinct elements in the list.
This is my code, but it is not working correctly. Please help! Thanks!!
(define (count-dist-elements lis)
(cond
((null? lis) 0)
((null? (cdr lis))0)
((member (car lis)(cdr lis)))
(else(+ 1(count-dist-elements (cdr lis))))))
p/s: let it be (count-dist-elements '(1 2 1 1 2 3 4 5 5 6 6 7 7 8 8 8 9))
回答1:
It looks like you're getting pretty close.
- What happens when you pass your function a list with one element? What should your function return in this case?
- What about a two-element list with the same element (eg.
(5 5))? Does your function return a sensible value?
回答2:
First: Why are you returning zero in the (null? (cdr lis)) case?
Second: What do you think your code returns in the case where the first element also occurs later in the list? Are you sure?
回答3:
(define (count-dist-elements lst dist-elems count)
(cond ((null? lst) count)
((member (car lst) dist-elems)
(count-dist-elements (cdr lst) dist-elems count))
(else
(count-dist-elements (cdr lst)
(cons (car lst) dist-elems)
(+ 1 count)))))
(count-dist-elements '(a b b c) '() 0) ==> 3
(count-dist-elements '(1 2 1 1 2 3 4 5 5 6 6 7 7 8 8 8 9) '() 0) ==> 9
Or, if you want it recursive and not iterative (and, it has to use a function call like the one shown),
(define (count-dist-elements lst . dist-elems)
(let ((dist-elems (if (null? dist-elems) '() (car dist-elems))))
(cond ((null? lst) 0)
((member (car lst) dist-elems)
(count-dist-elements (cdr lst) dist-elems))
(else
(+ 1 (count-dist-elements (cdr lst) (cons (car lst) dist-elems)))))))
Gives the same results.
(count-dist-elements '(1 2 1 1 2 3 4 5 5 6 6 7 7 8 8 8 9)) ==> 9
来源：https://stackoverflow.com/questions/5226526/counting-unique-elements-in-scheme