linked-list

Java two equal signs in one statement? [duplicate]

由 ぐ巨炮叔叔 提交于 2019-12-04 03:30:00
问题 This question already has answers here : Java - Order of Operations - Using Two Assignment Operators in a Single Line (4 answers) Closed 5 years ago . Can someone help me understand what the following code does and what the line with two equal sign does? How does something equal to something equal to something work in this constructor? public More ...LinkedList() { header.next = header.previous = header; } Here is the link to the website where I saw this and I'm trying to figure it out: http:

difference between double-ended linked lists and doubly-linked list

由 眉间皱痕 提交于 2019-12-04 02:36:39
I don't understand difference between a double-ended and doubly-linked list. What is the major difference between the two? In a doubly linked list, each node has two pointers. One towards its next node and another one towards its previous node. In a double-ended linked list, each node has just one pointer which points to its next node. Its difference from the single-ended linked list is that instead of just one "head" node, it contains two pointers of this kind ("first" and "last"), so someone is able to insert elements to list from both ends of it. (Last picture isn't that clear, but it

debug help - swap 2 nodes of double link list

由 ￣綄美尐妖づ 提交于 2019-12-04 02:27:53
问题 Could you please help me debug this code to swap two node of double link list? I am not able to figure out what i am doing wrong :( here is the code: dll* swap_node(dll *head , dll *node1 , dll *node2) { dll *tmp; int flag=0; if(node1->prev!=NULL) { node1->prev->next=node2; } else { flag=1; } if(node1->next!=NULL) { node1->next->prev=node2; } if(node2->prev!=NULL) { node2->prev->next=node1; } if(node2->next!=NULL) { node2->next->prev=node1; } tmp=node1->next; node1->next=node2->next; node2-

Why LinkedList and arraylist extends AbstractList in java?

由 此生再无相见时 提交于 2019-12-04 02:26:42
Why LinkedList and ArrayList extends AbstractList in Java ? Abstract classes are used when we want to specify a common behaviour in implementation classes. But all the methods which are in AbstractList are overridden by ArrayList and LinkedList . So what is the use of extending this class? sanbhat subList(int,int) method is not overriden by both ArrayList and LinkedList , and for this AbstractList provides a common implementation From Java source public List<E> subList(int fromIndex, int toIndex) { return (this instanceof RandomAccess ? new RandomAccessSubList<E>(this, fromIndex, toIndex) :

Declaring a LinkedList in Java

由 会有一股神秘感。 提交于 2019-12-04 01:57:14
I always learn when we declare a collection we should do, Interface ob = new Class() , if i want to use for example a LinkedList i'll do List ob = new LinkedList() , but then i can't have access to all methods from LinkedList.. Isn't LinkedList ob = new LinkedList() 100% correct? Isn't LinkedList ob = new LinkedList() 100% correct? Well I'd suggest using the generic form, but sure - if you want to use functionality which is specific to LinkedList , you need to declare the variable accordingly. You might want to check whether the Deque<E> or Queue<E> interfaces have what you want though. If

complexity of mergesort with linked list

由 痴心易碎 提交于 2019-12-04 01:52:42
问题 i have code for mergesort using linked list,it works fine,my question what is complexity of this algorithm?is it O(nlog(n))?also is it stable?i am interested because as i know mergesort is stable,what about using linked list?if we have elements with some equal with each-other,does this code preserve orders of elements?thanks a lot #include<stdio.h> #include <stdlib.h> struct node { int number; struct node *next; }; struct node *addnode(int number,struct node *next); struct node*mergesort

Is Java's LinkedList optimized to do get(index) in reverse when necessary?

由 笑着哭i 提交于 2019-12-04 01:26:46
问题 I've been working on some ways to optimize LinkedList's. Does anyone know if the Java default doubly-linked LinkedList class is optimized to do get() operations in reverse? For example: // Some LinkedList list that exists with n elements; int half = list.size() / 2; list.get(half + 1); Would the call to list.get(half + 1) optimize the search and go in reverse since it is a doubly-linked list? It would make more sense to do the search from the end and go towards the center if you know the

Difference between arraylist and linkedList [duplicate]

由 拥有回忆 提交于 2019-12-04 01:16:39
This question already has answers here : Closed 7 years ago . Possible Duplicate: When to use LinkedList<> over ArrayList<>? When to use a linked list over an array/array list? When should I use arrayList and when should I go for LinkedList? When should I use TreeSet , LinkedHashSet and HashSet ? When should i use arrayList and when should I go for LinkedList? Arraylist maintain indices like arrays. So if want more frequent get operations than put then arraylist is best to go. LinkedList maintain pointers to elements. you can't to a specific index like in arraylist. But the advantage here in

What is the Definition of a Lisp Cons Cell?

由 喜你入骨 提交于 2019-12-04 00:43:14
What exactly is the definition of a Common Lisp Cons Cell? How is a Cons Cell different than a standard linked list item? After all, both the cons cell and the linked list item have a value and a pointer to the next cell or item... or is this understanding wrong? Cons cells in general hold two pointers that can point to anything. General usage of course is to point to a "value" with the left one, and to another Cons cell (or nil) with the "right" one. A cons cell is closer to a binary tree node than a linked list node. car and cdr return the two children, which can be nil, atoms, or other cons

Why are Scala's `Lists` implemented as linked lists

由 孤街醉人 提交于 2019-12-04 00:18:31
I always thought that the benefit of linked lists was that you could add or remove items (especially not from the end) without having to copy lots of elements thanks to the beauty of pointers. However, Scala's List is immutable (at least by default). What is the benefit of having an immutable linked list (because there are definite downsides, e.g. not O(1) element access.) Thanks! I think the main reason is that one of the most powerful uses of linked lists is head/tail splitting. There are lots of recursive algorithms that look like this one: def listlen[A](xs: List[A], already: Int = 0): Int
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