Geek asks and answers

Reverse List - Scheme

I am trying to change a list, here is my code:

(define (reverse list)
  (if (null? list) 
     list
      (list (reverse (cdr list)) (car list))))

so if I enter (reverse) (1 2 3 4)), I want it to appear as (4 3 2 1), but right now this is not giving me this. What am I doing wrong and how can I fix it?

+5
source share
8 answers

- . append, , @lancery, - , , , , , - list , , . , , lst.

, consing , - , . , :

(define (reverse lst)
  (<???> lst '()))                       ; call the helper procedure

(define (reverse-aux lst acc)
  (if <???>                              ; if the list is empty
      <???>                              ; return the accumulator
      (reverse-aux <???>                 ; advance the recursion over the list
                   (cons <???> <???>)))) ; cons current element with accumulator

, reverse , .

+12

let:

(define (reverse lst)
  (let loop ([lst lst] [lst-reversed '()])
    (if (empty? lst)
        lst-reversed
        (loop (rest lst) (cons (first lst) lst-reversed)))))

, , , loop let , .

+1

, ( )

(define (reverse lst)
  (define (go lst tail)
    (if (null? lst) tail
        (go (cdr lst) (cons (car lst) tail))))
  (go lst ())))

(reverse ( 1 2 3 4))

;; (reverse (list 1 2 3 4))                                                                                                                           
;; (go (list 1 2 3 4) ())                                                                                                                             
;; (go (list 2 3 4) (list 1))                                                                                                                         
;; (go (list 3 4) (list 2 1))                                                                                                                         
;; (go (list 4) (list 3 2 1))                                                                                                                         
;; (go () (list 4 3 2 1))                                                                                                                             
;; (list 4 3 2 1)

, ( )

(define (reverse2 lst)
  (if (null? lst) ()
    (append (reverse2 (cdr lst)) (list (car lst)))))

(define (append l1 l2)
  (if (null? l1) l2
      (cons (car l1) (append (cdr l1) l2))))

(reverse2 ( 1 2 3 4))

;; (reverse2 (list 1 2 3 4))                                                                                                                          
;; (append (reverse2 (list 2 3 4)) (list 1))                                                                                                          
;; (append (append (reverse2 (list 3 4)) (list 2)) (list 1))                                                                                          
;; (append (append (append (reverse2 (list 4)) (list 3)) (list 2)) (list 1))                                                                          
;; (append (append (append (append (reverse2 ()) (list 4)) (list 3)) (list 2)) (list 1))                                                              
;; (append (append (append (append () (list 4)) (list 3)) (list 2)) (list 1))                                                                         
;; (append (append (append (list 4) (list 3)) (list 2)) (list 1))                                                                                     
;; (append (append (list 4 3) (list 2)) (list 1))                                                                                                     
;; (append (list 4 3 2) (list 1))                                                                                                                     
;; (list 4 3 2 1)
+1

build-list:

(define reverse
  (lambda (l)
    (let ((len (length l)))
      (build-list len
                  (lambda (i)
                    (list-ref l (- len i 1)))))))
0

, :

(define (rev lst)
 (if (null? lst)
     '()
      (append (rev (cdr lst)) (car lst))))
0

I think it is better to use append instead of cons

(define (myrev l)
  (if (null? l)
      '()
      (append (myrev (cdr l)) (list (car l)))
  )
)

this other version with tail recursion

(define (myrev2 l)
  (define (loop l acc) 
    (if (null? l)
        acc
        (loop (cdr l) (append (list (car l)) acc ))
    )
  )
  (loop l '())
)
0
source 
(define reverse?
  (lambda (l)
    (define reverse-aux?
      (lambda (l col)
        (cond 
          ((null? l) (col ))
          (else 
            (reverse-aux? (cdr l) 
                          (lambda () 
                            (cons (car l) (col))))))))
    (reverse-aux? l (lambda () (quote ())))))
(reverse? '(1 2 3 4) )

Another answer similar to an Oscar. I just started to study the circuit, so excuse me if you find problems :).

-1
source

There is really no need to add or fill the body with a bunch of lambda.

(define (reverse items)
  (if (null? items)
      '()
      (cons (reverse (cdr items)) (car items))))
-1
source

More articles:


All Articles