linked-list

问题 I'm working on a linux project. I need to pass a linked list to the kernel from a userspace program. I have used the kernel way of implementing linked lists in userspace. I have defined a structure as follows: struct time{ int val; struct list_head* list; }; The variable for this strucure would be: struct time mylist; I have given the input data and used list_add function found in linux/include/list.h to link the structures. This mylist is a part of another structure: struct params{ int data1

问题 I'm working on a linux project. I need to pass a linked list to the kernel from a userspace program. I have used the kernel way of implementing linked lists in userspace. I have defined a structure as follows: struct time{ int val; struct list_head* list; }; The variable for this strucure would be: struct time mylist; I have given the input data and used list_add function found in linux/include/list.h to link the structures. This mylist is a part of another structure: struct params{ int data1

问题 I'm working on a linux project. I need to pass a linked list to the kernel from a userspace program. I have used the kernel way of implementing linked lists in userspace. I have defined a structure as follows: struct time{ int val; struct list_head* list; }; The variable for this strucure would be: struct time mylist; I have given the input data and used list_add function found in linux/include/list.h to link the structures. This mylist is a part of another structure: struct params{ int data1

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 Most of sites present this solution(see Method 3 at http://www.geeksforgeeks.org/write-a-function-to-get-the-intersection-point-of-two-linked-lists/ ) 1) Get count of the nodes in first list, let count be c1. 2) Get count of the nodes in second list, let count be c2. 3) Get the difference of counts d = abs(c1 – c2) 4) Now traverse the bigger list from the first node till d nodes so that from here onwards both the lists have equal no of nodes. 5) Then we can traverse both the lists in parallel

问题 Most of sites present this solution(see Method 3 at http://www.geeksforgeeks.org/write-a-function-to-get-the-intersection-point-of-two-linked-lists/ ) 1) Get count of the nodes in first list, let count be c1. 2) Get count of the nodes in second list, let count be c2. 3) Get the difference of counts d = abs(c1 – c2) 4) Now traverse the bigger list from the first node till d nodes so that from here onwards both the lists have equal no of nodes. 5) Then we can traverse both the lists in parallel

问题 Most of sites present this solution(see Method 3 at http://www.geeksforgeeks.org/write-a-function-to-get-the-intersection-point-of-two-linked-lists/ ) 1) Get count of the nodes in first list, let count be c1. 2) Get count of the nodes in second list, let count be c2. 3) Get the difference of counts d = abs(c1 – c2) 4) Now traverse the bigger list from the first node till d nodes so that from here onwards both the lists have equal no of nodes. 5) Then we can traverse both the lists in parallel

问题 This question already has answers here : Closed 9 years ago . Possible Duplicate: When should I use a List vs a LinkedList If I expect not to use the access by index for my data structure how much do I save by using LinkedList over List ? If if am not 100% sure I will never use access by index, I would like to know the difference. Suppose I have N instances. inserting and removing in a LinkedList will only be a o(1) op , where as in List it may me be O(n), but since it it optimized, it would

问题 #include <iostream> #include <string> using namespace std; template <class T> class Node{ friend class LinkedList<T>; private: T data; Node <T> *next; public: Node(); Node(T d); ~Node(); }; template <class T> Node<T>::Node(){ T data = 0; next = 0; } template <class T> Node<T>::Node(T d){ data = d; next = 0; } template<class T> Node<T>::~Node(){ delete next; } template <class T> class LinkedList{ private: Node <T> *head; public: LinkedList(); ~LinkedList(); void Push_Front(const T& e); }

问题 I'm trying to move items in my list but when I compare against the last option I exit out before I move the items in my move linked list. Is there a way to do that before the node gets put at the end and can't loop through to move the items? LinkedList<BD> list = new LinkedList<BD>(b[arg].Values); LinkedListNode<BD> node, terminator, next = null; List<LinkedListNode<BD>> move = new List<LinkedListNode<BD>>(); terminator = list.First; node = next = list.Last; while (next != null && next !=