如果它们在Scheme中彼此相邻，则从列表中删除多个字符 [英] Remove multiple characters from a list if they are next to each other in Scheme

问题描述

我必须制作一个Dr.Racket程序，如果它们跟自己相同的字母，则从列表中移除字母。例如：（z z f a b b d d）将成为
（z f a b d）。我已经编写了代码，但它所做的只是删除列表中的第一个字母。
任何人都可以帮助？

I have to make a Dr. Racket program that removes letters from a list if they are following the same letter as itself. For example: (z z f a b b d d) would become (z f a b d). I have written code for this but all it does is remove the first letter from the list. Can anyone help?

#lang racket 
(define (remove-duplicates x)
(cond ((null? x)
     '())
    ((member (car x) (cons(car(cdr x)) '())))
     (remove-duplicates (cdr x))
    (else
     (cons (car x) (remove-duplicates (cdr x))))))

(define x '( b c c d d a a))
(remove-duplicates x)

推荐答案

(define (remove-dups x)
   (cond
     [(empty? x) '()]
     [(empty? (cdr x))  (list (car x))]
     [(eq? (car x) (cadr x))  (remove-dups (cdr x))]
     [else  (cons (car x) (remove-dups (cdr x)))]))

（cadr x） / code>是（car（cdr x））的缩写，以防您不知道。

(cadr x) is short for (car (cdr x)) in case you didn't know.

此外，模式匹配使列表解构通常更易读。在这种情况下，这不是很多，但是还是比其他版本还要好：

Also, pattern matching makes list deconstruction often much more readable. In this case not so much, but it's still better than the other version:

(define (rmv-dups x)
  (match x
    [(list)  (list)]
    [(list a)  (list a)]
    [(cons a (cons a b))  (rmv-dups (cdr x))]
    [__  (cons (car x) (rmv-dups (cdr x)))]))

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