Structure and Interpretation os Computer Programs Exercises Answer
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(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))))) |
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(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<br>(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 ‘()))) |
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：
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。
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(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))) |
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(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)) |
(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。
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(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))))) |
略
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(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)))))))))))))))))) |
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- (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)) |
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(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)))])) |
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(define (fringe items) (if
(null? items) ‘() (if
(pair? (car items)) (append
(fringe (car items)) (fringe (cdr items))) (cons (car items) (fringe (cdr items)))))) |
a)
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(define (left-branch bran) (car bran))(define (right-branch) (car (cdr bran))) |
1)不使用高阶函数，直接定义
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(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
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(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)))))) |
答案同2.30
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(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)) |
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(define (horner-eval x coefficient-sequence) (accumulate (lambda (this-coeff higher-terms) (+ this-coeff (* x higher-terms))) 0 coefficient-sequence)) |
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(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))))) |
原文：http://www.cnblogs.com/richard-g/p/3589813.html