Question

我有这段代码：

(define remove
  (λ (x y)
    (cond 
          ((null? y) '())
          ((eq? x (car y)) (delete x (cdr y)))
          (else (cons (car y) (delete x (cdr y))))
)))

Input: (remove 'c '(w((x)(c q)(((o))))w))

这不会进入内括号内。如果我尝试删除＆＃39; w＆＃39;，它将删除所有出现的内容，因为它们在括号外。所以，我无法从内部取出任何东西。

Answer 1

解决方案稍微复杂一点，您必须使用该模板遍历列表列表。此外，最好使用equal?代替eq?，请参阅documentation以了解差异。接下来是一个可能的实现，我冒昧地将参数重命名为更有意义的东西：

(define remove
  (λ (ele lst)
    (cond
      ((null? lst) '())           ; if the list is empty then we're done
      ((not (pair? (car lst)))    ; if the first element is an atom
       (if (equal? (car lst) ele) ; check if it's the one we're looking for
           (remove ele (cdr lst)) ; if so we skip over it, eliminating it
           (cons (car lst) (remove ele (cdr lst))))) ; otherwise we add it
      (else (cons (remove ele (car lst))      ; else advance recursion
                  (remove ele (cdr lst))))))) ; over both `car` and `cdr`

现在它按预期工作：

(remove 'c '(w ((x) (c q) (((o)))) w))
=> '(w ((x) (q) (((o)))) w)

取出列表中所有元素的出现

1 个答案: