Question

我必须制作Dr. Racket程序，如果它们跟随自己的相同字母，则从列表中删除字母。例如：（z z f a b b d d）将成为（z f a b d）。我已经为此编写了代码，但它只是从列表中删除了第一个字母。有人可以帮忙吗？

#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)

Answer 1

(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)))]))

如果您不知道，

(cadr x)是(car (cdr x))的缩写。

此外，模式匹配使列表解构通常更具可读性。在这种情况下，没有那么多，但它仍然比其他版本更好：

(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)))]))

Answer 2

如果引入辅助函数，这个问题会更简单。

我推荐这样的东西（尖括号意味着你需要填写细节）：

(define (remove-duplicates x)
  (cond
    [ <x is empty>                             '()]  ; no duplicates in empty list
    [ <x has one element>                      x]    ; no duplicates in a list with one element
    [ <first and second element in x is equal> (cons (car x) (remove-from-front (car x) (cdr x)))]
    [else                                      (cons (car x) (remove-duplicates (cdr x)))]))

(define (remove-from-front e x)
  (cond
    [ <x is empty>                  '()]                 ; e is not the first element of x
    [ <e equals first element of x> (remove-from-front e (cdr x))] ; skip duplicate
    [else                           (remove-duplicates x)]))       ; no more es to remove

如果它们在Scheme中彼此相邻，则从列表中删除多个字符

2 个答案: