SICP习题解答2.17-2.23

#lang racket

; exercise 2.17
(define (last-pair l)
(cond ((null? l) null)
((null? (cdr l)) l)
(else (last-pair (cdr l)))))
(last-pair (list 23 72 149 34))
(last-pair '(1))
(last-pair '())

; exercise 2.18
(define (reverse l)
(define (first-n-1 l)
(if (null? (cdr l))
null
(cons (car l) (first-n-1 (cdr l)))))
(if (null? l)
null
(cons (car (last-pair l)) (reverse (first-n-1 l)))))
(reverse (list 1 4 9 16 25)) ;; '(25 16 9 4 1)

; exercise 2.19
(define (cc amount coin-values)
(cond ((= amount 0) 1)
((or (< amount 0) (no-more? coin-values)) 0)
(else (+ (cc amount (except-first-denomination coin-values))
(cc (- amount (first-denomination coin-values)) coin-values)))))
(define (no-more? l)
(null? l))
(define (except-first-denomination l)
(cdr l))
(define (first-denomination l)
(car l))
(define us-coins (list 50 25 10 5 1))
(define uk-coins (list 100 50 20 10 5 2 1 0.5))
(cc 100 us-coins) ;; 292
;(cc 100 uk-coins) ;; 104561
;; coins-value的排列不会影响cc的结果

; exercise 2.20
(define (same-parity x . y)
(define (sp x y)
(define (odd? n) (= (remainder n 2) 1))
(cond ((null? y) null)
((odd? (+ x (car y))) (sp x (cdr y)))
(else (cons (car y) (sp x (cdr y))))))
(cons x (sp x y)))
(same-parity 1 2 3 4 5 6 7)
(same-parity 2 3 4 5 6 7)

; exercise 2.21
(define (square-list items)
(if (null? items)
null
(cons (square (car items)) (square-list (cdr items)))))
(define (square-list2 items)
(map square items))
(define (square x) (* x x))
(square-list (list 1 2 3 4))
(square-list2 (list 1 2 3 4))

; exercise 2.23
(define (for-each proc l)
(cond ((not (null? l))
(proc (car l))
(for-each proc (cdr l)))))
(for-each (lambda (x) (newline) (display x))
(list 57 321 88))