在方案中，如何将元素添加到列表中一定次数？

Question

so the function would take 3 elements a number item and a list. I want to add the item to the list how ever many times the number says so if i did (pad-front 3 'a '(bc)) it would return (aaabc) as the list. I think i would want to cons the item to the list and just do that n times im just not sure on how to get it to do it.

Answer 1

In Racket this is easy using built-in procedures:

(define (pad-front n x lst)
  (append                   ; append together both lists
   (build-list n (const x)) ; create a list with n repetitions of x
   lst))                    ; the list at the tail

(pad-front 3 'a '(b c))
=> '(a a a b c)

... But I guess you want to implement it from scratch. The idea is the same as above, but using only primitive procedures; I'll give you some hints so you can figure out the details. Fill-in the blanks:

(define (pad-front n x lst)
  (if <???>            ; base case: if `n` is zero
      <???>            ; then return the tail list
      (cons <???>      ; else cons the element to be repeated
            (pad-front ; and advance the recursion
             <???>     ; subtract one unit from `n`
             x lst)))) ; pass along `x` and `lst`

Notice that the trick is to keep adding the x element until no more repetitions are needed (in other words: until n is zero). At that point, we just stick the tail list at the end, and we're done!