circular-list

问题 I am trying to find an algorithm to count the number of nodes in a Circular linked list by using one pointer only . Does anyone know any algorithm? 回答1: Try checking this link: https://www.geeksforgeeks.org/count-nodes-circular-linked-list/ We can also use array to keep count of the nodes visited and exit one's the node count becomes 2. But this approach works for linked list with unique elements only. 来源： https://stackoverflow.com/questions/55487427/circular-linked-list-count-number-of-nodes

问题 Is there a standard function in Common Lisp that can check against improper lists (i.e. circular and dotted lists) without signaling an error? list-length can check against circular lists (it returns nil for them), but signals type-error when given a dotted list. Scheme's list? traverses the whole list to make sure it is not dotted or circular; Common Lisp's listp only checks that it's given nil or a cons cell. Here's the simplest I could come up with: (defun proper-list-p (x) (not (null

问题 Is there a standard function in Common Lisp that can check against improper lists (i.e. circular and dotted lists) without signaling an error? list-length can check against circular lists (it returns nil for them), but signals type-error when given a dotted list. Scheme's list? traverses the whole list to make sure it is not dotted or circular; Common Lisp's listp only checks that it's given nil or a cons cell. Here's the simplest I could come up with: (defun proper-list-p (x) (not (null

问题 My question was marked as duplicate, and closed How to deal with cycles in recursive functions in scheme? I'm asking again: I have this code: (define x (list 1 2 3)) (set-cdr! (cddr x) x) (define (list? x) (and (pair? x) (or (null? (cdr x)) (list? (cdr x))))) (display (list? x)) (newline) and function list? freezes when it find cycle (because it's infinite loop). How scheme code usually work with cycles like that? I'm not asking how to detect cycles. I'm asking: how this is done in scheme

问题 Ok so the idea for me is to be moving to each node(user in this case) in a circular list and asking if they would like to log off,they will give a random yes or no answer,until everyone has logged off. this seems to be the case most of the time i run the program but sometimes users are logging back on which shouldn't happen,I will post the delete method and the display method i am using. public void displayLinkedList() { temp=first; int i = 1; do { boolean rand=randomBoolean(); if(rand) {

问题 Here is my attempt: case class A(val a: A, val b: Int){ override def toString() = b.toString } lazy val x: A = A(y, 0) lazy val y: A = A(z, 1) lazy val z: A = A(x, 2) The problem comes when trying to do anything with x; causing x to be evaluated starts off a circular evaluation going through x, y, z and ends in a stack overflow. Is there a way of specifying that val a should be computed lazily? 回答1: You need to make A.a itself lazy. You can do it by turning it into a by name parameter that is

问题 I am working using a visual programming environment for musical composition based on CL . I am trying to create a function that when given say 3 elements (1 2 3) will return 1, 2, 3, 1, 2, 3 etc., one number at the time each time it is evaluated. The book Common Lisp a Gentle Introduction , mentions briefly that it's possible to create circular lists using sharp-equal notation but does not get into details on how to use them. Keep in mind that I can insert actual Lisp code in the program

问题 In Scheme/Lisp I am trying to make a function that converts a list into a circular list. Therefore, I believe I need to construct an infinite stream in which the tail of the list points to the head of the list. Here is my code so far: (define (rotate-list l1 l1copy) (if (null? (force (cdr l1))) (cons (car l1) (delay l1copy))) (cons (car l1) (delay (rotate-list (force (cdr l1)) l1copy)))) All help is greatly appreciated. 回答1: No, you don't need streams to make a circular list. There are two

问题 Closed . This question needs to be more focused. It is not currently accepting answers. Want to improve this question? Update the question so it focuses on one problem only by editing this post. Closed 12 months ago . How would I go about iterating through a list from 1-100, where I delete every other element starting with the first element, and repeat that step until there in only one element left in the list. Would I have to use a circular linked list, or can it be done just using loops and

问题 Referencing @dfeuer's answer to this question: Least expensive way to construct cyclic list in Haskell, which says that using cyclic lists 'defeats' the garbage collector as it has to keep everything you've consumed from a cyclic list allocated till you drop the reference to any cons cells in the list. Apparently in Haskell a cyclic list and an infinite list are two separate things. This blog (https://unspecified.wordpress.com/2010/03/30/a-doubly-linked-list-in-haskell/) says that if you