Le 03/06/2015 07:25, Paul Bian a écrit :
Hi all,
one function to remove duplicates from the left,
i.e. 1 2 1 3 2 4 5 -> 1 2 3 4 5
and one from the right.
i.e. 1 2 1 3 2 4 5 -> 1 3 2 4 5
Hello,
For the first function, I think you can't do it without a local define
or by adding a second parameter (for result). Here is my solution :
(define (right L)
(cond ((empty? L) (list))
((member? (car L) (cdr L)) (right (cdr L)))
(else (cons (car L) (right (cdr L))))))
(define (left L res)
(cond ((empty? L) res)
((not (member? (car L) res)) (left (cdr L) (append res
(list(car L)))))
(else (left (cdr L) res))))
For left, I use an empty list at initialisation (res), and I can check
if an element have already been added. I didn't use reverse but it is
recommanded instead of append cause it cost less of computing time.
append = reverse = O(n) . But i use append more than one time.
In practise I never use remove, cause it cost a lot of computing time
too O(pos) (cause have to reindex each element of the list each time),
except if you remove the first element, but (cdr) does it ;) .
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