气泡分选方案 [英] Bubble Sorting in Scheme

问题描述

我正在将递归代码编写为Bubble Sort(通过交换从最小到最大)
我有一个代码可以只对气泡进行一次排序

I am writing a recursive code to Bubble Sort (smallest to largest by swapping)
I have a code to do the bubble sort just once

(define (bubble-up L)  
   (if (null? (cdr L))  
     L   
  (if (< (car L) (cadr L))  
(cons (car L) (bubble-up (cdr L)))  
(cons (cadr L) (bubble-up (cons (car L) (cddr L))))  
  )
 )  

如果我在此代码中放入了一个列表，它将返回结尾处编号最大的列表.
EX ..(冒泡'(8 9 4 2 6 7))->'(8 4 2 6 7 9)

if i put a list into this code, it returns the list with the largest number at the end
EX.. (bubble-up ' (8 9 4 2 6 7)) -> ' (8 4 2 6 7 9)

现在我正在尝试编写代码以执行(泡沫L)N次(列表中整数的数量)
我有以下代码:

Now i am trying to write a code to do the (bubble-up L) N times (the number of integers in list)
I have this code:

  (define (bubble-sort-aux N L)   
    (cond ((= N 1) (bubble-up L))  
       (else (bubble-sort-aux (- N 1) L)  
  (bubble-up L))))  
(bubble-sort-aux 6 (list 8 9 4 2 6 7))  -> ' (8 4 2 6 7 9)

但是递归似乎没有发生，因为它只排序一次！
任何建议都将受到欢迎，我很困惑！

But the recursion doesn't seem to happen because it only sorts once!
Any suggestions would be welcome, i'm just stumped!

推荐答案

尝试一下:

(define (bubble-sort-aux N L)   
  (cond ((= N 1) (bubble-up L))  
        (else (bubble-sort-aux (- N 1) (bubble-up L)))))  

如果您继续冒泡"列表N次，它将在末尾排序.您的代码的问题在于您没有将bubble-up的结果用于任何东西-但是，如果我们将bubble-up返回的值传递给函数的下一个调用，则最终将对它进行排序.现在，该过程将按预期工作:

If you keep "bubbling-up" the list N times it'll be sorted at the end. The problem with your code is that you weren't using the result of bubble-up for anything - but if we pass the value returned by bubble-up to the next call of the function, it'll eventually be sorted. Now the procedure works as expected:

(bubble-sort-aux 6 (list 8 9 4 2 6 7))
=> '(2 4 6 7 8 9)

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