具有一個參數的方案功能，它將每兩個元素交換

Question

(define (interchange list) 
    (if (empty? list) 
     list 
     (interchange (append (car (cdr list) X))))) 

我需要創建一個函數來交換方案列表中的元素對。這就是我來了這麼遠，但我得到的錯誤與empty?

Error 
empty?: undefined; 
cannot reference undefined identifier 

function call          output 
(interchange '())         () 
(interchange '(a))         (a) 
(interchange '(a b))        (b a) 
(interchange '(a b c))       (b a c) 
(interchange '(a 1 b 2 c 3 d 4))    (1 a 2 b 3 c 4 d) 
(interchange '(hello you -12.34 5 -6 enough)) (you hello 5 -12.34 enough -6) 

Answer 1

試試這個：

(define (interchange lst) 
    (if (or (null? lst) (null? (cdr lst)))  ; if the list has 0 or 1 elements left 
     lst          ; then return the list, we're done 
     (cons (cadr lst)       ; otherwise grab the second element 
      (cons (car lst)     ; and the first element 
        (interchange (cddr lst)))))) ; and move forward two elements 

關鍵是要處理每次迭代的兩個元素，注意與邊緣案件。它的工作原理如預期的樣本輸入：

(interchange '()) 
=> '() 
(interchange '(a)) 
=> '(a) 
(interchange '(a b)) 
=> '(b a) 
(interchange '(a b c)) 
=> '(b a c) 
(interchange '(a 1 b 2 c 3 d 4)) 
=> '(1 a 2 b 3 c 4 d) 
(interchange '(hello you -12.34 5 -6 enough)) 
=> '(you hello 5 -12.34 enough -6) 

Answer 2

#lang racket 

(define (interchange xs) 
    (match xs 
    [(or '() (list _))     ; if the list is empty or has one element 
    xs]        ; then interchange keeps the original list 
    [(list* x y more)     ; if the list has two or more elements, 
    (list* y x (interchange more))])) ; then the result consists of the two 
;          ; first elements (with the order swapped) 
;          ; followed by the result of 
;          ; interchanging the remaining elements. 


(interchange '())         
(interchange '(a))        
(interchange '(a b))        
(interchange '(a b c))       
(interchange '(a 1 b 2 c 3 d 4))    
(interchange '(hello you -12.34 5 -6 enough)) 

輸出：

'() 
'(a) 
'(b a) 
'(b a c) 
'(1 a 2 b 3 c 4 d) 
'(you hello 5 -12.34 enough -6)