Question

我想知道如何用另一个符号替换列表中的符号。这是我想出的，

;; swap:  s1 s2 los -> los 
;; given a list of symbols and symbol1 symbol2 
;; return a list with all occurrences of s1 replaced with 
;; s2 and all occurrences of s2 replaced with s1

(define (swap s1 s2 1st)
   (cond 
      [(empty? 1st) empty]
      [(cons? 1st)
          (cond
              [(symbol=? (first 1st) s1) (swap s2 s1 (rest 1st))]
              [else (cons (first 1st)
                    (swap  s1 s2 (rest 1st)))])]))

测试;

（交换＆＃39; a＆＃39; d（＆＃39; a＆＃39; b＆＃39; c＆＃39; d））=＆gt;列表（＆＃39; d＆＃39; b＆＃39; c＆＃39; a）

好吧，看起来我的代码只是摆脱它们而不是互相替换它们。有什么建议在这里出了什么问题吗？

我想的可能，

[(symbol=? (first 1st) s1) (swap s2 s1 (rest 1st))]

应重写为

[(symbol=? (first 1st) s1) (cons s2 (rest 1st))]

这有助于将＆＃39; a替换为＆＃39;

但是如何在else递归过程中替换＆＃39; a？

Answer 1

你很亲密，但如果找到cons，你忘了s1：

(define (swap s1 s2 1st)
  (cond 
    [(empty? 1st) empty]
    [(cons? 1st)
     (cond
       [(symbol=? (first 1st) s1) (cons s2 (swap s2 s1 (rest 1st)))] ; <--- cons added
       [else (cons (first 1st)
                   (swap s1 s2 (rest 1st)))])]))

测试：

> (swap 'a 'd (list 'a 'b 'c 'd))
'(d b c a)

scheme用另一个符号替换列表中的符号

1 个答案: