计划功能以反向列表 [英] Scheme function to reverse a list

问题描述

对于我的编程语言类，我应该在Scheme中编写一个函数以反转列表，而无需使用预制的反转函数.到目前为止，我得到的是

For my programming languages class I'm supposed to write a function in Scheme to reverse a list without using the pre-made reverse function. So far what I got was

(define (reverseList lst)
(COND
((NULL? lst) '())
(ELSE (CONS (reverseList(CDR lst)) (CAR lst)))
))

我遇到的问题是，如果我输入列表，可以说(a b c)它给了我(((() . c) . b) . a).

The problem I'm having is that if I input a list, lets say (a b c) it gives me (((() . c) . b) . a).

我应该如何在没有多套括号和.的情况下获得一个干净的列表?

How am I supposed to get a clean list without multiple sets of parenthesis and the .'s?

推荐答案

实现的问题是cons没有收到 list 作为第二个参数，所以答案是重建不是正确的列表，请记住:要正确构建列表，是通过cons将元素与列表一起构建，而最后一个列表为空.

The problem with your implementation is that cons isn't receiving a list as its second parameter, so the answer you're building isn't a proper list, remember: a proper list is constructed by consing an element with a list, and the last list is empty.

一个可能的解决方法是使用一个辅助函数，该函数在累加器参数中构建答案，cons反向添加元素-顺便说一下，此解决方案是尾递归:

One possible workaround for this is to use a helper function that builds the answer in an accumulator parameter, consing the elements in reverse - incidentally, this solution is tail recursive:

(define (reverse lst)
  (reverse-helper lst '()))

(define (reverse-helper lst acc)
  (if (null? lst)
      acc
      (reverse-helper (cdr lst) (cons (car lst) acc))))

(reverse '(1 2 3 4 5))
=> '(5 4 3 2 1)

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