SICP_2.61-2.62

#lang racket

;;;;;;;;;;;;;;;;;2.61
(define (element-of-set? x set)
  (cond ((null? set) false)
  ((= x (car set)) true)
  ((< x (car set)) false)
  (else (element-of-set? x (cdr set)))))

(define (intersection-set set1 set2)
  (if (or (null? set1) (null? set2))
      '()
      (let ((x1 (car set1)) (x2 (car set2)))
        (cond ((= x1 x2)
               (cons x1
                     (intersection-set (cdr set1)
                                      (cdr set2))))
              ((< x1 x2)
               (intersection-set (cdr set1) set2))
              ((< x2 x1)
               (intersection-set set1 (cdr set2)))))))

(define (adjion-set x set)
  (cond 
        ((null? set) (cons x set))
        ((= x (car set)) set)
        ((< x (car set)) (cons x set))
        ((> x (car set)) (cons (car set) (adjion-set x (cdr set))))))

;;;;;;;;;;;;;;;;;;;2.62 只有当两列表有顺序时成立
(define (union-set set1 set2)
  (cond ((null? set1) set2)
        ((null? set2) set1)
        ((< (car set1) (car set2))
         (cons (car set1) (union-set (cdr set1) set2)))
        ((< (car set2) (car set1))
         (cons (car set2) (union-set set1 (cdr set2))))
        ((= (car set1) (car set2))
         (cons (car set1) (union-set (cdr set1) (cdr set2))))))

;;;;;;;;;;;;test
(adjion-set 6 '(2 3 4 5 7))
(adjion-set 1 '(2 3 4 5 7))
(adjion-set 8 '(2 3 4 5 7))
(union-set '(1 2 3 4 5) '(2 3 4 5 7))
(union-set '(1 3 5 6 8) '(2 3 4 5 9))

注意2.62的算法是在有顺序条件下才能实现的