Question

(define-struct school (name students))
;; An SchoolChart is a (make-school Str (listof SchoolChart))
;; names are unique

我有学校图表说

(define s-chart (make-school "Tom" (list 
(make-school "James" empty) 
(make-school "Claire" 
    (make-school "David" empty) 
    (make-school "Travis" empty)) 
(make-school "Timmy" empty))))

这是一般树，比如说我定义了一个函数

(define (find-name name school)) ;;produces true if found/false if not.

我如何进行递归？这个具体案例很好，但每个孩子都可以有无限的孩子吗？我只需要一个提示

Answer 1

只能有一定数量的孩子数量是任意的，只受机器内存的限制，但不能无限。

（而你的public ActionResult TransferDevices(IList<XXXViewModel> viewModel) { var selected = viewModel.Where(x => x.isSelected).ToList(); IQueryable<Device> devicesQueryable = _db.Devices; var devices = devicesQueryable.Where(x => selected.Any(y => y.Id == x.Id)).ToList() ; }格式不正确，因为“克莱尔”的孩子不是名单。）

递归可以非常简单这是深度优先搜索：

s-chart

其中(define (find-name name school) (or (string=? name (school-name school)) (any (lambda (s) (find-name name s)) (school-students school))))为(any p ls)当且仅当#t为(p e)时，列表#t的至少一个元素e。< / p>

现在剩下的就是写ls ...

Answer 2

以递归方式检查所有项目，如果找到，则将名称添加到循环外的列表中。但是，它需要使用set!。它使用string-prefix?代替string=?用于演示目的（在当前结构中获取更多名称）：

(define-struct school (name students))

(define s-chart
  (make-school "Tom"
               (list 
                (make-school "James" empty) 
                (make-school "Claire" (list
                                       (make-school "David" empty) 
                                       (make-school "Travis" empty)))
                (make-school "Timmy" empty))))


(define (find-name name school)
  (define ol '())
  (let loop ((s school))
    (cond
      [(list? s)
       (when (not(empty? s))
           (begin (loop (first s))
                  (loop (rest s))))]
      [else
       (when (string-prefix? (school-name s) name)
         (set! ol (cons (school-name s) ol)))
       (loop (school-students s))
         ]))
  ol
  )

(find-name "T" s-chart)

输出：

'("Timmy" "Travis" "Tom")

在嵌套列表Scheme / Racket中搜索

2 个答案: