# 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

• 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。

    (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

### 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2.25

    (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2.33

    (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2.36

    (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)))))


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posted @ 2014-03-09 12:21  Liqiang Gao  阅读(256)  评论(0编辑  收藏  举报