Racket - 在二维列表中查找具有相同元素的列表

Question

有人可以告诉我如何解决这个问题/给我一个模板吗？我不允许使用 map 或 lambda 或任何其他高级内置函数，只能列出

Answer 1

首先提取与输入字符串关联的国家：

(define (get-country l s)
  (cond [(empty? (rest l)) (second (second (first l)))]
        [else (if (equal? s (first (first l)))
                  (second (second (first l)))
                  (get-country (rest l) s))]))

然后提取与该国家/地区相关联的所有字符串：

(define (get-countries l s)
  (cond [(empty? l) '()]
        [else (if (equal? s (second (second (first l))))
                  (cons (first (first l)) (get-countries (rest l) s))
                  (get-countries (rest l) s))]))

然后把它们放在一起：

(define (same-country l s)
  (get-countries l (get-country l s)))

当我们评估时，我们得到的结果与(list "YUL" "YVR" "YWG" "YYZ") 不同：

> (same-country alist "YUL")
(list "YYZ" "YWG" "YUL" "YVR")

所以我们检查结果是否是所需列表的排列。首先我们制作is-permutation:

(define (is-permutation l1 l2)
  (and (not (and (cons? l1) (empty? l2)))
       (not (and (empty? l1) (cons? l2)))
       (or (and (empty? l1) (empty? l2))
           (and (is-member (first l1) l2)
                (is-permutation (rest l1)
                                (remove-one (first l1) l2))))))

(define (is-member e l)
  (and (not (empty? l))
       (or (equal? (first l) e)
           (is-member e (rest l)))))

(define (remove-one e nel)
  (cond [(empty? (rest nel)) '()]
        [else (if (equal? (first nel) e)
                  (rest nel)
                  (cons (first nel) (remove-one e (rest nel))))]))

然后我们可以测试：

> (is-permutation (same-country alist "YUL")
                  (list "YUL" "YVR" "YWG" "YYZ"))
#true