SICP 习题2.59 union-set操作

union-set操作的想法就是找出set1中set2里面没有的，set2中没有的和set2放在一起，就是union-set

第二种实现方法利用了adjoin,其实思想是一样的，

（union set1 set2） = (union (cdr set1) (adjoin (car set1) set2))

(define (element-of-set? x set)
  (cond ((null? set) false)
        ((equal? x (car set)) true)
        (else (element-of-set? x (cdr set)))))

(define (adjoin x set)
  (if (element-of-set? x set)
    set
    (cons x set)))

(define (intersection-set set1 set2)
  (cond ((or (null? set1) (null? set2)) '())
         ((element-of-set? (car set1) set2)
          (cons (car set1)
                (intersection-set (cdr set1) set2)))
         (else (intersection-set (cdr set1) set2))))

(define (union-set set1 set2)
  (cond ((null? set1) set2)
        ((not (element-of-set? (car set1) set2))
         (cons (car set1)
            (union-set (cdr set1) set2)))
        (else (union-set (cdr set1) set2))))

(define s1 '(1 2 3 4 5 6))
(define s2 '(9 2 3 4 8 10))

(newline)
(display (intersection-set s1 s2))

(newline)
(display (union-set s1 s2))

(define (union-set set1 set2)
  (cond ((null? set1) set2)
        (else 
          (union-set (cdr set1)
                     (adjoin (car set1) set2)))))

(newline)
(display (union-set s1 s2))