我正在尝试创建一个方案尾递归函数 flatten-tl-rec 来展平嵌套列表列表。
(define flatten-tl-rec
(lambda (xs)
(letrec ([flatten-tl-rec-acc
(lambda (xs acc)
(cond ((empty? xs) acc)
((list? (first xs)) (flatten-tl-rec-acc (rest xs) (append (flatten-tl-rec-acc (first xs) '()) acc)))
(else (flatten-tl-rec-acc (rest xs) (cons (first xs) acc))))
)])
(flatten-tl-rec-acc xs '()))))
(flatten-tl-rec '(1 2 3 (4 5 6) ((7 8 9) 10 (11 (12 13)))))
但我越来越(13 12 11 10 9 8 7 6 5 4 3 2 1)
代替(1 2 3 4 5 6 7 8 9 10 11 12 13)
。
这里出了什么问题?
您正在列表错误的末尾累积元素。您可以将它们附加到列表的正确末尾:
(define flatten-tl-rec
(lambda (xs)
(letrec ([flatten-tl-rec-acc
(lambda (xs acc)
(cond ((empty? xs) acc)
((list? (first xs))
(flatten-tl-rec-acc
(rest xs)
(append acc (flatten-tl-rec-acc (first xs) '()))))
(else (flatten-tl-rec-acc
(rest xs)
(append acc (list (first xs)))))))])
(flatten-tl-rec-acc xs '()))))
...或者简单地反转最后的列表:
(define flatten-tl-rec
(lambda (xs)
(letrec ([flatten-tl-rec-acc
(lambda (xs acc)
(cond ((empty? xs) acc)
((list? (first xs))
(flatten-tl-rec-acc
(rest xs)
(append (flatten-tl-rec-acc (first xs) '()) acc)))
(else (flatten-tl-rec-acc
(rest xs)
(cons (first xs) acc)))))])
(reverse (flatten-tl-rec-acc xs '())))))
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