1 #lang racket
2
3 ;;;;;;;;;;;;;;;;;
2.61
4 (define (element-of-
set? x
set)
5 (cond ((
null?
set)
false)
6 ((= x (car
set))
true)
7 ((< x (car
set))
false)
8 (
else (element-of-
set? x (cdr
set)))))
9
10 (define (intersection-
set set1 set2)
11 (
if (or (
null? set1) (
null?
set2))
12 '()
13 (let ((x1 (car set1)) (x2 (car set2)))
14 (cond ((=
x1 x2)
15 (cons x1
16 (intersection-
set (cdr set1)
17 (cdr set2))))
18 ((<
x1 x2)
19 (intersection-
set (cdr set1) set2))
20 ((<
x2 x1)
21 (intersection-
set set1 (cdr set2)))))))
22
23 (define (adjion-
set x
set)
24 (cond
25 ((
null?
set) (cons x
set))
26 ((= x (car
set))
set)
27 ((< x (car
set)) (cons x
set))
28 ((> x (car
set)) (cons (car
set) (adjion-
set x (cdr
set))))))
29
30 ;;;;;;;;;;;;;;;;;;;
2.62 只有当两列表有顺序时成立
31 (define (union-
set set1 set2)
32 (cond ((
null?
set1) set2)
33 ((
null?
set2) set1)
34 ((<
(car set1) (car set2))
35 (cons (car set1) (union-
set (cdr set1) set2)))
36 ((<
(car set2) (car set1))
37 (cons (car set2) (union-
set set1 (cdr set2))))
38 ((=
(car set1) (car set2))
39 (cons (car set1) (union-
set (cdr set1) (cdr set2))))))
40
41 ;;;;;;;;;;;;test
42 (adjion-
set 6 '(2 3 4 5 7))
43 (adjion-
set 1 '(2 3 4 5 7))
44 (adjion-
set 8 '(2 3 4 5 7))
45 (union-
set '(1 2 3 4 5) '(
2 3 4 5 7))
46 (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
转载于:https://www.cnblogs.com/tclan126/p/6481461.html
相关资源：数据结构—成绩单生成器
