CODEWITHSUNDEEP
recursionscheme
sarahka7
sarahka7

Reputation: 23

Finding duplicate elements in a Scheme List

I'm trying to become more familiar with recursion in Scheme. I have searched the question bank and see many "built-in" functions for finding duplicate entries in a scheme list, but am trying to design my own. I have not learned about "lambda" yet either. My concern is when I call the recursion function, the car element will be updated. I wish to keep it the same, but pass a new cdr each time, so the original car element can be compared against each subsequent element in the list. I want to return a #t if there is a match, and #f if there is no match or the cdr is empty (base case). Any help/advice would be greatly appreciated. 

(define (findDuplicates list)
    (if (null? list)
       #f
       (if (null? (cdr list))
           #f
             (if (= (car list) (getCarOfCdr list)) 
              #t
              (findDuplicates (cdr list))    //trying to use recursion
             )
        )
    )
)


(define (getCarOfCdr list)          //Helper function
  (car (cdr list))
)

Upvotes: 0

Views: 2841

Answers (1)

Óscar López
Óscar López

Reputation: 235984

You should use at least one built-in procedure to make things simpler: member (which is standard). Some general suggestions:

  • You should not call a parameter list, that clashes with a built-in function with the same name.
  • Don't nest ifs like that, use cond instead.
  • We only need two base cases for this problem, not three as in your code.
  • getCarOfCdr is not necessary: it can be replaced by a call to cadr and besides, that won't work: you have to check each element against all the rest of the list - that's why you should use member.

Taking all of the above suggestions into consideration, here's my proposed solution:

(define (findDuplicates lst)
  (cond ((null? lst) #f)
        ((member (car lst) (cdr lst)) #t)
        (else (findDuplicates (cdr lst)))))

If using member doesn't satisfy your requirements, then it's easy to implement, and simpler than findDuplicates; you should try to write your own version, just for fun.

Upvotes: 3

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