list - 方案满屋戳功能

Question

我正在尝试在 Scheme 中定义一个函数来确定一个五元素列表是否包含满屋(即分别有 3 个元素相同和另外 2 个元素相同)。我脑子里有大纲，虽然我弄乱了语法。我正在使用 and, let, to try and do this.输入是一个列表(5 个元素，编号 1-13)，输出是一个 bool 值。这是我目前所拥有的:

(define is-full-house?
  (lambda (listy)
    ;; Sort listy from smallest to greatest
    (let ((sorted-list (sort listy <=)))
        (and 
         ((= (first sorted-list) (second sorted-list)) (= (fourth sorted-list) (fifth sorted-list))))
        (or
         ((= third fourth)) (= first third)))))

感谢帮助

Answer 1

您很接近 - 但请确保您首先了解 满屋，以及在这种情况下使用 bool 连接符的正确方法。试试这个:

(define is-full-house?
  (lambda (listy)
    (let ((sorted-list (sort listy <=)))
      (or
       (and
        (= (first sorted-list) (second sorted-list) (third sorted-list))
        (= (fourth sorted-list) (fifth sorted-list)))
       (and
        (= (first sorted-list) (second sorted-list))
        (= (third sorted-list) (fourth sorted-list) (fifth sorted-list)))))))