計(jì)算機(jī)程序的構(gòu)造和解釋習(xí)題解答

Structure and Interpretation os Computer Programs Exercises Answer

第二章 構(gòu)造數(shù)據(jù)抽象

練習(xí)2.17

    (define last-pair-1

      (lambda (input)

        (if (= (length input) 1)

            input

            (last-pair-1 (cdr input)))))

	;因?yàn)閘ength的定義是遞歸定義的，所以如果是一個(gè)長(zhǎng)列表，用length會(huì)非常耗時(shí)

	;last-pair-2和last-pair-3用null？去檢測(cè)列表是否為空，效率會(huì)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)))))

練習(xí)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 '())))

練習(xí)2.19

理解換零錢的邏輯，將總數(shù)為a的現(xiàn)金換成n種硬幣的不同方式等于：

• 不考慮第一種硬幣，將現(xiàn)金a換成除第一種硬幣之外的所有其他硬幣的不同方式數(shù)目，加上
• 考慮第一種硬幣，將現(xiàn)金a-v[0]換成所有種類的硬幣的不同方式的數(shù)目。

所以表coin-values的排列順序給結(jié)果沒有關(guān)系，是否有序排列結(jié)果都是一樣的。

    (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)))

練習(xí)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))

練習(xí)2.22

(cons x y)的作用是把x當(dāng)成一個(gè)元素插入到列表y的開頭，如果x本身是一

個(gè)列表，x會(huì)以列表的身份插入到y(tǒng)開頭。

比如(cons '(1) '(2 3))的結(jié)果不是'(1 2 3)，而是'('(1) 2 3)。

此處可以使用append。

練習(xí)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)))))

練習(xí)2.24

略

練習(xí)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))))))))))))))))))

練習(xí)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))

練習(xí)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)))]))

練習(xí)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))))))

練習(xí)2.29

a)

    (define (left-branch bran)

      (car bran))

    (define (right-branch)

      (car (cdr bran)))

練習(xí)2.30

1)不使用高階函數(shù)，直接定義

    (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))))))

練習(xí)2.31

答案同2.30

練習(xí)2.32

練習(xí)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))

練習(xí)2.34

    (define (horner-eval x coefficient-sequence)

      (accumulate (lambda (this-coeff higher-terms) (+ this-coeff (* x higher-terms)))

                  0

                  coefficient-sequence))

練習(xí)2.35

練習(xí)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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