3

我需要在 Scheme lang的列表中的输入索引上切换 2 个元素。例如：

(交换索引 0 3 '(1 2 3 4 5))

（4 2 3 1 5）

有人可以帮忙吗？提前致谢！:)

4

3 回答 3

1

现在我想不出一种方法来解决这个问题，除非在整个列表中迭代三遍，（每个一个list-ref，一个多一个build-list。）不是最有效的解决方案，但它是这样的：

(define (swap-index idx1 idx2 lst)
  (define (build-list lst idx e1 e2)
    (cond ((null? lst)
           '())
          ((= idx idx1)
           (cons e2 (build-list (cdr lst) (add1 idx) e1 e2)))
          ((= idx idx2)
           (cons e1 (build-list (cdr lst) (add1 idx) e1 e2)))
          (else
           (cons (car lst) (build-list (cdr lst) (add1 idx) e1 e2)))))
  (build-list lst 0 (list-ref lst idx1) (list-ref lst idx2)))

我假设给定列表的索引存在，否则list-ref会产生错误。索引可以按任何顺序传递，这意味着：idx1可以小于、等于或大于idx2。它按预期工作，返回一个带有修改的新列表：

(swap-index 0 3 '(1 2 3 4 5))
=> '(4 2 3 1 5)
于 2012-11-26T13:48:06.560 回答
1

此方法最多遍历列表两次：

(define (swap-index index1 index2 lst)
    ;; FIND-ELEMENTS -- 
    ;;  INPUT:  count, an integer; lst, a list
    ;; OUTPUT:  a pair of the form '(a . b)
    (define (find-elements count lst)
      (cond ((null? lst) '()) ; really, we should never reach this if indices are valid
            ((= count index1) ; found the first element, so hold on to it while we look for the next one
             (cons (car lst) (find-elements (+ 1 count) (cdr lst))))
            ((= count index2) (car lst)) ; found the second element, return part 2 of the pair
            (else ; since we only care about 2 elements we can just skip everything else
              (find-elements (+ 1 count) (cdr lst)))))
    ;; BUILD-LIST --
    ;;  INPUT:  count, an integer; elements, a pair; lst, a list
    ;; OUTPUT:  a new list
    (define (build-list count elements lst)
      (cond ((null? lst) '()) ; again, we shouldn't get here if indices are valid
            ((= count index1) ; reached first index, substitute 2nd element and keep going
             (cons (cdr elements) (build-list (+ 1 count) elements (cdr lst))))
            ((= count index2) ; reached second index, substitute 1st element and stop
             (cons (car elements) (cdr lst)))
            (else  ; everything else just gets added to the list per usual
              (cons (car lst) (build-list (+ 1 count) elements (cdr lst))))))
    (build-list 0 (find-elements 0 lst) lst)) ; call build-list using a call to find-elements as a parameter

首先，find-elements查看列表并返回一cons对我们想要交换的元素。 注意： 此代码取决于按顺序给出索引的假设，以便最小的在前。

接下来，build-list获取输出，find-elements以便在下一次遍历期间我们可以替换适当的元素。

于 2012-11-26T23:21:26.333 回答
0

这是clojure中的解决方案。我希望算法会有所帮助。

(defn split [idx lst]
  (let [lst-rest (drop idx lst)]
   [(take idx lst) (first lst-rest) (rest lst-rest)]))

(defn swap-index [idx1 idx2 lst]
  (let [[lst1 e1 lst] (split idx1 lst)
        [lst2 e2 lst3] (split (dec (- idx2 idx1)) lst)]
    (concat lst1 [e2] lst2 [e1] lst3)))

=> (swap-index 0 3 [1 2 3 4 5])
   (4 2 3 1 5)
于 2012-11-27T04:35:08.170 回答