如何使用方案删除列表中的所有重复项(仅允许抽象列表函数) [英] How to remove all the duplicates in a list using scheme (only abstract list functions allowed)

问题描述

我知道如何以递归方式编写此代码.

I know how to write this in an recursive way.

(define (removed2 lst)
  (cond
       [(empty? lst) empty]
       [(not (member? (first lst) (rest lst)))
        (cons (first lst) (removed2 (rest lst)))]
       [else (removed2 (rest lst))]))

so (removed2 (list 1 1 1 2 2 2 3 3 3 3 3 3)) 给出 (list 1 2 3)

so (removed2 (list 1 1 1 2 2 2 3 3 3 3 3 3)) gives (list 1 2 3)

但是，如何仅使用抽象函数(filter、foldr、map 和 build-list)来重写它?

However, how do you rewrite it only using abstract functions (filter, foldr, map, and build-list)?

我尝试使用过滤器，但它不起作用.

I tried to use filter but it just doesn't work.

推荐答案

可以使用 foldr 来解决这个问题，诀窍是知道在 lambda 中写什么:

It's possible to use foldr for this, the trick is knowing what to write in the lambda:

(define (removed lst)
  (foldr (lambda (e a)
           (if (not (member? e a))
               (cons e a)
               a))
         '()
         lst))

还要检查您的解释器是否具有内置功能，例如在 Racket 中您可以使用 remove-duplicates.

Also check if your interpreter has a built-in function, for instance in Racket you can use remove-duplicates.

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