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练习2.17
;; 直接利用已经实现的list-ref和length过程即可(define (last-pair items) (if (null? items) (display "null") (list-ref items (- (length items) 1))))
练习2.18
;; 翻转即为将列表第二个元素翻转再加上第一个元素(define (reverse items) (if (null? items) '() (append (reverse (cdr items)) (list (car items)))));; 翻转即为倒序取列表值再组合成新的列表(define (reverse items) (define (reverse-iter a n) (if (< n 0) '() (cons (list-ref a n) (reverse-iter a (- n 1))))) (reverse-iter items (- (length items) 1)))
练习2.19
(define (no-more? coin-values) (null? coin-values))(define (except-first-denomination coin-values) (cdr coin-values))(define (first-denomination coin-values) (car coin-values)) 1 ]=> (cc 100 us-coins);Value: 2921 ]=> (cc 100 (reverse us-coins));Value: 292;; 改变coin-values的顺序不会影响结果;; 因为在cc过程中递归的累加 只用某种币值的兑换种数和除去这种币值后的兑换种数, 因此和次数无关
练习2.20
;; 首先定义flag过程, 以判断传入的两个数是否奇偶性一致;; 然后在get-list过程中递归使用cdr取列表的剩余部分完成奇偶性检查;; 最后将参数拼接成列表调用get-list过程(define (same-parity x . y) (define (flag a b) (= (remainder a 2) (remainder b 2))) (define (get-list items) (if (null? items) '() (if (flag x (car items)) (cons (car items) (get-list (cdr items))) (get-list (cdr items))))) (get-list (cons x y))) 1 ]=> (same-parity 1 2 3 4 5 6 7);Value : (1 3 5 7)1 ]=> (same-parity 2 3 4 5 6 7);Value : (2 4 6)1 ]=> (same-parity 2 2 0 128 6 7);Value : (2 2 0 128 6)
练习2.21
(define (square-list items) (if (null? items) '() (cons (square (car items)) (square-list (cdr items))))) (define (square-list items) (map (lambda (x) (square x)) items))
练习2.22
;; Louis在组成新的结果时, 在cons的两个参数上搞反了;; 调整之后因为对列表和数值直接结合导致出现如下结果;Value : (((((() . 1) . 4) . 9) . 16) . 25);; 因此做如下修正, 将数值变换为列表然后再和之前的结果相加(define (square-list items) (define (iter things answer) (if (null? things) answer (iter (cdr things) (append answer (list (square (car things))))))) (iter items '()))
练习2.23
(define (for-each proc items) (if (null? items) '() (and (proc (car items)) (for-each proc (cdr items)))))
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