String 从scheme中的字符串中删除重复字符

这个问题我已经试了很长时间了，但还没有完全解决。问题是生成一个字符串，其中输入字符串中的所有重复字符都被该字符的单个实例替换

比如说,

(remove-repeats "aaaab") => "ab"
(remove-repeats "caaabb aa") => "cab a"

由于我试图使用累积递归来实现这一点，到目前为止，我已经：

(define (remove-repeats s) 
  (local
    [(define (remove-repeats-acc s1 removed-so-far)
      (cond
        [(empty? (string->list s1))""]
        [else 
         (cond
           [(equal? (first (string->list s1)) (second (string->list s1))) 
         (list->string (remove-repeats-acc (remove (second (string->list s1)) (string->list s1)) (add1 removed-so-far)))]
           [else (list->string (remove-repeats-acc (rest (string->list s1)) removed-so-far))])]))]
    (remove-repeats-acc s 0)))

但这似乎不对。请帮我修改这个以使其工作

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谢谢

字符串使用起来有点烦人，因此我们将其包装在处理列表的辅助函数中。这样我们就可以避免到处乱搞转换

(define (remove-repeats str)
  (list->string (remove-repeats/list (string->list str))))

现在，我们可以使用简单的递归定义remove repeats/list函数：

(define (remove-repeats/list xs)
  (cond
    [(empty? xs) xs]
    [(empty? (cdr xs)) xs]
    [(equal? (car xs) (cadr xs)) (remove-repeats/list (cdr xs))]
    [else (cons (car xs) (remove-repeats/list (cdr xs)))]))

这不是尾部递归，但现在应该更容易添加累加器：

(define (remove-repeats str)
  (list->string (remove-repeats/list-acc (string->list str) '())))

(define (remove-repeats/list-acc xs acc)
  (cond
    [(empty? xs) (reverse acc)]
    [(empty? (cdr xs)) (reverse (cons (car xs) acc))]
    [(equal? (car xs) (cadr xs)) (remove-repeats/list-acc (cdr xs) acc)]
    [else (remove-repeats/list-acc (cdr xs) (cons (car xs) acc))]))

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以下是我喜欢的一个版本：

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请将您的代码放在代码块中，适当缩进操作系统，这样更易于阅读。您可以通过在每行之前至少添加四个空格以及在适当的位置开始新行，使代码看起来更漂亮。谢谢！！真的很感激

#lang typed/racket
(: remove-repeats : String -> String)
(define (remove-repeats s)
  (define-values (chars last)
    (for/fold: ([chars : (Listof Char) null] [last : (Option Char) #f])
      ([c (in-string s)] #:when (not (eqv? last c)))
      (values (cons c chars) c)))
  (list->string (reverse chars)))