SICP 习题2.60 允许元素重复，重构集合操作

element-of-set? 和intersection-set的复杂度急剧上升，而adjoin-set和union-set的复杂度下降了一个$O(n)$。

在adjoin-set操作和union-set操作比较多，element-of-set?和intersection-set操作比较少的时候

(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) (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)
  (append set1 set2))

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

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

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