使用CPS删除Scheme中的子序列函数(深度递归) [英] Remove subsequence function (deep recursion) in Scheme using CPS

问题描述

removesub*包含一个原子列表和一个常规列表.第一个列表是第二个列表的子序列.该方法应返回第二个列表，并删除第一次出现的子序列.因此，如果第一个列表为'(abc)，则删除第二个列表的第一个a，删除在删除的a之后出现的第一个b，以及删除的b之后出现的第一个c-无论如何原子嵌套的深度.

removesub* takes a list of atoms and a general list. The first list is a subsequence of the second list. The method should return the second list with the first occurence of the subsequence removed. So, if the first list is '(a b c), the first a if the second list is removed, the first b that appears after the removed a is removed, and the first c that appears after the removed b is removed - no matter how deep the atoms are nested.

(removesub *'(a b)'(w(x b)((a)((y z)))b))

(removesub* '(a b) '(w (x b) ((a) ((y z))) b))

预期输出: (w(x b)(()((y z))))

Expected Output: (w (x b) (() ((y z))))

我正在尝试使用连续传递样式(CPS)来完成此功能.对于这种复杂的功能，我真的很难掌握. 借助先前的stackoverflow问题，我得以尝试问题，但我的尝试返回了一个空列表.

I am trying to complete this function using continuous passing style (CPS). This is really difficult for me to grasp with a function of this complexity. With the help of a previous stackoverflow question, I was able to attempt the problem, but my attempt returns an empty list.

我在做什么错了?

(define removesub*
  (lambda (l1 l2 return)
    (cond
      ((or (not (pair? l1)) (not (pair? l2))) return l1 l2)
      ((pair? (car l2))
       (removesub* l1
                   (car l2)
                   (lambda (v1 v2car)
                     (removesub* v1
                                 (cdr l2)
                                 (lambda (v1 v2cdr)
                                   (return v1 (cons v2car v2cdr)))))))
      ((eq? (car l1) (car l2))
       (removesub* (cdr l1) (cdr l2) return))
      (else
       (removesub* l1
                   (cdr l2)
                   (lambda (v1 v2)
                     (return v1 (con (car l2) v2))))))))

推荐答案

对您的代码进行了两处小的更改，我就能完成一些工作:

With two small changes to your code, I got something to work:

1. 我在第一个cond分支中将return l1 l2更改为(return l1 l2).
2. 我在底行将con更改为cons.

1. I changed return l1 l2 to (return l1 l2) in the first cond branch.
2. I changed con to cons in the bottom line.

祝你好运！

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