Three steps for deleting a node from a linked list:
1) set currentNode->prev->next to currentNode->next (i.e. the previous node's next pointer should be the current node's next pointer).
2) set currentNode->next->prev to currentNode->prev (i.e. the next node's previous pointer should be the current node's previous pointer).
3) Free the memory used by currentNode (using delete, for example).
examples:- delete this node (identified by a pointer)- insert a new node before this node- replace this node with another node
Lack of skill?
To delete an node in a linked list...If the head node is empty, stop, and return failed.Interate thorugh each member of the list.Remember at each iteration, the address of the privious node.If the current node is the head node or not the desired node, continue to the next iteration.Move the current node's pointer value to the previous node's pointer.Delete the current node, and return success.Return failed..
To delete a node (this) in a linked list, first you need to find the address of the parent node (parent).Iterate through the list, checking to find if the head pointer (head) or a child node (parent) points to (this).Store the next pointer of (this) in (parent) or (head), as determined by step 2.Delete (this).
If by 'head node' you simply mean the first node, then yes; but if 'head node' means the special element which is not supposed to ever be deleted (aka sentinel node), then no.
For a singly-linked list, only one pointer must be changed. If the node about to be deleted (let's call it node for the sake of argument) is the head of the list, then the head node pointer must be changed to node->next. Otherwise, the node that comes before the deleted node must change its next pointer to node->next. Note that given a singly-linked node has no knowledge of its previous node, we must traverse the list from the head in order to locate that particular node, unless the node is the head of the list: void remove (List* list, Node* node) { if (!list !node) return; // sanity check!if (list->head == node) {list->head = node->next;} else {Node* prev = list->head;while (prev->next != node) prev = prev->next; // locate the node's previous nodeprev->next = node->next;}} Note that the remove function only removes the node from the list, it does not delete it. This allows us to restore the node to its original position, because the node itself was never modified (and thus still refers to its next node in the list). So long as we restore all removed nodes in the reverse order they were removed, we can easily restore the list. In order to delete a node completely, we simply remove it and then free it:void delete (List* list, Node* node) {if (!list !node) return; // sanity check!remove (list, node);free (node);} For a doubly-linked list, either two or four pointers must be changed. If the node about to be deleted is the head node, then the head node pointer must be changed to n->next and n->next->prev must be changed to NULL, otherwise, n->prev->next becomes n->next. In addition, if the node about to be deleted is the tail node, then the tail node pointer must be changed to n->prev and n->prev->next must be changed to NULL, otherwise, n->next->prev becomes n->prev. Deletion from a doubly-linked list is generally quicker than deletion from a singly linked list because a node in a doubly-linked list knows both its previous node and its next node, so there's no need to traverse the list to locate the previous node to the one being deleted. void remove (List* list, Node* node) {if (!list !node) return; // sanity check!if (list->head == node) {list->head = node->next;node->next->prev = NULL;} else {node->prev->next = node->next; }if (list->tail == node) {list->tail = node->prev;node->prev->next = NULL;} else {node->next->prev = node->prev; }} Again, to physically delete the node we simply remove and then free the node:void delete (List* list, Node* node) {if (!list !node) return; // sanity check!remove (list, node); free (node); }
Yes. The tail node's next node is the head node, while the head node's previous node is the tail node.
I'm currently doing the same homework problem in c++, my best guess right now is to use some function such as max element size or sizeof(object) and then if and then statements or a for loop to go through the list to delete the smallest linked node.
Answersingly linked list has the node inserted only at one end. and the pointer corresponds to the next pointer.but in a doubly linked list, the node pointer points to the both previous and the next node.singly linked list has two nodesdoubly linked list has three nodesA doubly linked list makes sense when you need to traverse the list in both directions. You aren't able to do that with a singly linked list.
Yes, each node in a doubly linked list contain a link to the previous as well as the next node. That is the definition of the doubly linked list.
struct LinkedListNode { void* data; LinkedListNode* next; }; LinkedListNode* head; LinkedListNode* tmp; while (head) { tmp head->next; free(head); head tmp; }
The difference is how many pointers each node has, and what they are pointing to. A linked list is comprised of "Nodes" each node contains data as well as 1 or more pointers. A singly linked list has one pointer per node, and a doubly linked list has 2 pointers per node. Some programs use several pointers per node. The purpose of these pointers is to hold the list together. In a singly linked list, you can view a node and can then move on to the next node that it is pointing to until you've passed through them all. A doubly-linked list would have a pointer to the next node as well as to the previous node. Thus you can move forward and backward through the list. A circularly-linked list doesn't necessarily have a set number of pointers because it simply means that the last node points to the first node creating a big circle. A non-circularly-linked list would not contain this last to first pointer and thus you would eventually reach the end of the list and stop.