SICP 习题解 第二章
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原文链接：http://www.cnblogs.com/richard-g/p/3589813.html
计算机程序的构造和解释习题解答
Structure and Interpretation os Computer Programs Exercises Answer
第二章 构造数据抽象
练习2.17
(define last-pair-1
(lambda (input)
(if (= (length input) 1)
input
(last-pair-1 (cdr input)))))
;因为length的定义是递归定义的，所以如果是一个长列表，用length会非常耗时
;last-pair-2和last-pair-3用null？去检测列表是否为空，效率会更高
(define last-pair-2
(lambda (input)
(if (null? (cdr input))
input
(last-pair-2 (cdr input)))))
(define last-pair-3
(lambda (input)
(let ([last (cdr input)])
(if (null? last)
input
(last-pair-3 last)))))练习2.18
(define reverse-1
(lambda (input)
(if (null? input)
'()
(append (reverse-1 (cdr input)) (cons (car input) '())))))
;用迭代的方式效率更高
(define reverse-2
(lambda (input)
(define reverse-iter
(lambda (remain record)
(if (null? remain)
record
(reverse-iter (cdr remain) (cons (car remain) record)))))
(reverse-iter input '())))练习2.19
理解换零钱的逻辑，将总数为a的现金换成n种硬币的不同方式等于：
- 不考虑第一种硬币，将现金a换成除第一种硬币之外的所有其他硬币的不同方式数目，加上
- 考虑第一种硬币，将现金a-v[0]换成所有种类的硬币的不同方式的数目。
所以表coin-values的排列顺序给结果没有关系，是否有序排列结果都是一样的。
(define first-denomination
(lambda (coin-values)
(car coin-values)))
(define except-first-denomination
(lambda (coin-values)
(cdr coin-values)))
(define no-more?
(lambda (coin-values)
(if (null? coin-values)
#t
#f)))练习2.21
(define (square-list-1 items)
(if (null? items)
'()
(cons (* (car items) (car items)) (square-list-1 (cdr items)))))
(define (square-list-2 items)
(map (lambda (x) (* x x)) items))练习2.22
(cons x y)的作用是把x当成一个元素插入到列表y的开头，如果x本身是一
个列表，x会以列表的身份插入到y开头。
比如(cons '(1) '(2 3))的结果不是'(1 2 3)，而是'('(1) 2 3)。
此处可以使用append。
练习2.23
(define (for-each-1 func items)
(if (null? items)
#f
(or (func (car items))
(for-each-1 func (cdr items))))) (define (for-each-1 func items)
(if (null? items)
#f
(or (func (car items))
(for-each-1 func (cdr items)))))练习2.24
略
练习2.25
(car (cdr (car (cdr (cdr '(1 3 (5 7 9)))))))
(car (car '((7))))
(car (cdr (car (cdr (car (cdr (car(cdr (car (cdr (car (cdr '(1 (2 (3 (4 (5 (6 7))))))))))))))))))练习2.26
- (append x y): '(1 2 3 4 5 6)
- (cons x y): '((1 2 3) (4 5 6))
- (list x y): '((1 2 3) (4 5 6))练习2.27
(define (deep-reverse items)
(cond
[(null? items) (list)]
[(not (pair? (car items))) (append (deep-reverse (cdr items)) (list (car items)))]
[else (append (deep-reverse (cdr items)) (cons (deep-reverse (car items)) (list)))]))练习2.28
(define (fringe items)
(if (null? items)
'()
(if (pair? (car items))
(append (fringe (car items)) (fringe (cdr items)))
(cons (car items) (fringe (cdr items))))))练习2.29
a)
(define (left-branch bran)
(car bran))
(define (right-branch)
(car (cdr bran)))练习2.30
1)不使用高阶函数，直接定义
(define (square-tree items)
(if (null? items)
(list)
(if (not (pair? (car items)))
(cons (* (car items) (car items)) (square-tree (cdr items)))
(append (cons (square-tree (car items)) (list)) (square-tree (cdr items))))))2)使用map
(define (tree-map proc trees)
(if (null? trees)
(list)
(if (not (pair? (car trees)))
(cons (proc (car trees)) (tree-map proc (cdr trees)))
(append (cons (tree-map proc (car trees)) (list)) (tree-map proc (cdr trees))))))练习2.31
答案同2.30
练习2.32
练习2.33
(define (map p sequence)
(accumulate (lambda (x y) (p x)) '() sequence))
(define (append seq1 seq2)
(accumulate cons seq2 seq1))
(define (length sequence)
(accumulate (lambda (x y) (+ y 1)) 0 sequence))练习2.34
(define (horner-eval x coefficient-sequence)
(accumulate (lambda (this-coeff higher-terms) (+ this-coeff (* x higher-terms)))
0
coefficient-sequence))练习2.35
练习2.36
(define (first-elems items)
(if (null? items)
(list)
(cons (car (car items)) (first-elems (cdr items)))))
(define (rest-elems items)
(if (null? items)
(list)
(cons (cdr (car items)) (rest-elems (cdr items)))))
(define (accumulate-n op init seqs)
(if (null? (car seqs))
(list)
(cons (accumulate op init (first-elems seqs))
(accumulate-n op init (rest-elems seqs)))))转载于:https://www.cnblogs.com/richard-g/p/3589813.html
标签：cons,items,SICP,cdr,car,input,习题,第二章,define 来源： https://blog.csdn.net/weixin_30299539/article/details/98198541