方案 - 列表的递归函数
我有这个程序：
(define scale-tree
(lambda (tree factor)
(map (lambda (sub-tree)
(if (list? sub-tree)
(scale-tree sub-tree factor)
(* sub-tree factor)))
tree)))
(scale-tree (list 1 (list 2 (list 3 4) 5) (list 6 7))
10)
这段代码是如何工作的？首先，我们将整个列表作为参数 (list 1 (list 2 (list 3 4) 5) (list 6 7))，并且在第一次调用中， (lambda ( sub-tree) 获取 (list 1 (list 2 (list 3 4) 5) (list 6 7)) 作为参数。树子树Factor) 和 (list 1 (list 2 (list 3 4) 5) (list 6 7)) 再次，
谢谢。
I have this program:
(define scale-tree
(lambda (tree factor)
(map (lambda (sub-tree)
(if (list? sub-tree)
(scale-tree sub-tree factor)
(* sub-tree factor)))
tree)))
(scale-tree (list 1 (list 2 (list 3 4) 5) (list 6 7))
10)
How does this code work? First, we give it the whole list as parameter (list 1 (list 2 (list 3 4) 5) (list 6 7)), and in the first call, the (lambda (sub-tree) gets the (list 1 (list 2 (list 3 4) 5) (list 6 7)) as parameter. For that, we call (scale-tree sub-tree factor) with (list 1 (list 2 (list 3 4) 5) (list 6 7)) again. When does the list reduce?
Thank you.
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记住
map的作用 - 它将函数应用于列表的每个元素。因此，在第一次调用时，此函数：应用于列表的元素：
1、(list 2 (list 3 4) 5)和（列表 6 7）。递归调用中依此类推。Remember what
mapdoes - it applies a function to every element of a list. So on the first call, this function:is being applied to the elements of your list:
1,(list 2 (list 3 4) 5), and(list 6 7). And so on in the recursive calls.