Reputation: 837
how to design a function that merge two lists into one list. the first element of first list will be the first element of the new list and the first element of the second list will be the second element of the new list (a,b,c,d,e,f) (g,h,i) will be (a,g,b,h,c,i,d,e,f,)
Upvotes: 5
Views: 22358
Reputation: 9414
This can be done using a simple condition.
(define (merge a b)
(cond ((null? a) b)
((null? b) a)
(else (cons (car a) (merge b (cdr a))))
))
Upvotes: 1
Reputation: 19
There is no need to check both lists: Here is a simple version:
(define (interleave lx lz)
(cond
[(empty? lx) lz]
[else (cons (first lx)(interleave lz (rest lx)))]))
(check-expect(interleave '() '())'())
(check-expect(interleave '(X X X X X X) '(O O O O O))
(list 'X 'O 'X 'O 'X 'O 'X 'O 'X 'O 'X))
(check-expect(interleave '(1 2 3) '(a b c d e f))
(list 1 'a 2 'b 3 'c 'd 'e 'f))
Upvotes: 0
Reputation: 236004
The procedure you're trying to implement is known as interleave
or merge
. Because this looks like homework, I can't leave you a straight answer, instead I'll point you in the right direction; fill-in the blanks:
(define (interleave lst1 lst2)
(cond ((null? lst1) ; If the first list is empty
<???>) ; ... return the second list.
((null? lst2) ; If the second list is empty
<???>) ; ... return the first list.
(else ; If both lists are non-empty
(cons (car lst1) ; ... cons the first element of the first list
<???>)))) ; ... make a recursively call, advancing over the first
; ... list, inverting the order used to pass the lists.
Upvotes: 8
Reputation: 10324
Here is a purely functional and recursive implementation in R6RS
(define (merge l1 l2)
(if (null? l1) l2
(if (null? l2) l1
(cons (car l1) (cons (car l2) (merge (cdr l1) (cdr l2)))))))
Upvotes: 13