试图创建一个检查列表排序的函数

Question

我正在尝试创建一个函数，如果列表是正确的，则返回true或false。我不知道如何使它是递归的。我一直收到最后一行的错误消息: define: expected只对函数体有一个表达式，但是找到了一个额外的部分

(check-expect (is-sorted? (list)) true)
(check-expect (is-sorted? (list "A")) true)
(check-expect (is-sorted? (list "B" "A")) false)
(check-expect (is-sorted? (list "A" "B")) true)
(check-expect (is-sorted? (list "A" "C" "B")) false)
(check-expect (is-sorted? (list "A" "B" "C")) true)

(define (is-sorted? lst)
  (cond
    ((empty-list? lst) true)
    ((= (length lst) 1) true) ;return true if there is only one element.
    ((string<? (first lst) second lst) true) ;If the first element is smaller than the second 
                                              ;return true.
    (else (is-sorted? (rest lst))))) ;do the above steps on the rest of the elements in the list.

Answer 1

请注意，您没有考虑过程何时应该返回false的情况，并且当您发现一个元素相对于下一个元素排序时，提前退出(您需要继续迭代！)这只是一场比赛，其他的呢？)解决方案很简单，只需否定这种情况的条件，并询问它是否没有排序返回false。如下所示：

(define empty-list? empty?)

(define (is-sorted? lst)
  (cond
    ((empty-list? lst) true)
    ((empty-list? (rest lst)) true)
    ((string>=? (first lst) (second lst)) false)
    (else (is-sorted? (rest lst)))))

它适用于您的测试用例：

(is-sorted? (list)) ; true
(is-sorted? (list "A")) ; true
(is-sorted? (list "B" "A")) ; false
(is-sorted? (list "A" "B")) ; true
(is-sorted? (list "A" "C" "B")) ; false
(is-sorted? (list "A" "B" "C")) ; true