<< Previous exercise (2.61) | Index | Next exercise (2.63) >>
Here's a slightly more elegant solution using ajoint-set from the previous exercise.
;From ex. 2.61 (define (adjoin-set x set) (cond ((null? set) (list x)) ((= x (car set)) set) ((> x (car set)) (cons (car set) (adjoin-set x (cdr set)))) ((< x (car set)) (cons x set)) (else (error "WTF!?")))) ;Answer (define (union-set s1 s2) (cond ((null? s1) s2) ((element-of-set? (car s1) s2) (union-set (cdr s1) s2)) (else (union-set (cdr s1) (adjoin-set (car s1) s2)))))
Exercise 2.62. Give a Θ(n) implementation of union-set for sets represented as ordered lists.
(define (union-set set1 set2) (cond ((null? set1) set2) ((null? set2) set1) ((= (car set1) (car set2)) (cons (car set1) (union-set (cdr set1) (cdr set2)))) ((< (car set1) (car set2)) (cons (car set1) (union-set (cdr set1) set2))) (else (cons (car set2) (union-set set1 (cdr set2))))))Or, If you want to make it ugly saving 5 car operations on a run,
(define (union-set set1 set2) (cond ((null? set1) set2) ((null? set2) set1) (else (let ((x1 (car set1)) (x2 (car set2))) (cond ((= x1 x2) (cons x1 (union-set (cdr set1) (cdr set2)))) ((< x1 x2) (cons x1 (union-set (cdr set1) set2))) (else (cons x2 (union-set set1 (cdr set2)))))))))