sicp每日一题[2.24-2.27]

2.24-2.26没什么代码量，所以跟 2.27 一起发吧。

Exercise 2.24

Suppose we evaluate the expression (list 1 (list 2 (list 3 4))). Give the result printed by the interpreter, the corresponding box-and-pointer structure, and the interpretation of this as a tree (as in Figure 2.6).

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解释器执行结果为：'(1 (2 (3 4)))
另外两种表示方法见下图：

Exercise 2.25

Give combinations of cars and cdrs that will pick 7 from each of the following lists:

(1 3 (5 7) 9)
((7))
(1 (2 (3 (4 (5 (6 7))))))

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纯看耐心。。

; 先创建这些要求的列表
(list 1 3 (list 5 7) 9)
(list (list 7))
(list 1 (list 2 (list 3 (list 4 (list 5 (list 6 7))))))

; 根据它们的结构依次定位到 7
(car (cdr (car (cdr (cdr (list 1 3 (list 5 7) 9))))))
(car (car (list (list 7))))
(car (cdr (car (cdr (car (cdr (car (cdr (car (cdr (car (cdr (list 1 (list 2 (list 3 (list 4 (list 5 (list 6 7))))))))))))))))))

; 执行结果
'(1 3 (5 7) 9)
'((7))
'(1 (2 (3 (4 (5 (6 7))))))
7
7
7

Exercise 2.26

Suppose we define x and y to be two lists:

(define x (list 1 2 3))
(define y (list 4 5 6))

What result is printed by the interpreter in response to evaluating each of the following expressions:

(append x y)
(cons x y)
(list x y)

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结果如下所示：

'(1 2 3 4 5 6)
'((1 2 3) 4 5 6)
'((1 2 3) (4 5 6))

Exercise 2.27

Modify your reverse procedure of Exercise 2.18 to produce a deep-reverse procedure that takes a list as argument and returns as its value the list with its elements reversed and with all sublists deep-reversed as well. For example,

(define x (list (list 1 2) (list 3 4)))
x
((1 2) (3 4))
(reverse x)
((3 4) (1 2))
(deep-reverse x)
((4 3) (2 1))

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这道题难度还是挺大的，我开始想用 reverse 实现，比起原来的 reverse 函数，deep-reverse 需要在 items 不为空的情况下，额外判断 (car items) 是否为 pair，虽然也能做，但是步骤有点麻烦，最后在网上发现一个使用 append 的版本，程序非常简单，而且逻辑也很清楚。

(define (deep-reverse-by-reverse items) 
  (define (iter items result) 
    (if (null? items) 
        result 
        (if (pair? (car items)) 
            (let ((x (iter (car items) nil))) 
              (iter (cdr items) (cons x result))) 
            (iter (cdr items) (cons (car items) result))))) 
  (iter items nil))

(define (deep-reverse items)
  (if (pair? items)
      (append (deep-reverse (cdr items))
              (list (deep-reverse (car items))))
      items))


(define x (list (list 1 2) (list 3 4)))

x
(reverse x)
(deep-reverse x)