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deleting a node
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Algoritm for deleting the last element from a list?
Given a list and a node to delete, use the following algorithm: // Are we deleting the head node? if (node == list.head) { // Yes -- assign its next node as the new head list.head = node.next } else // The node is not the head node { // Point to the head node prev = list.head // Traverse the list to locate the node that comes immediately before the one we want to delete while (prev.next != node) { prev = prev.next; } end while // Assign the node's next node to the previous node's next node prev.next = node.next; } end if // Before deleting the node, reset its next node node.next = null; // Now delete the node. delete node;
Is deleting the only node in linked list a special case?
It depends on whether or not you have a head node. If you do, then it is not a special case. If you don't, then it is, and you need to be able to update the head pointer.
What is time complexity to delete a node in singly linked list?
Time complexity to locate a node in a list of n nodes is O(1) at best (first node), O(n) at worst (last node) and O(n/2) on average. Once located, deleting a node takes constant time O(1).
Explain any two advantages using Single linked list over Doubly linked list and vice-versa?
Advantages of single linked list: # Decrease in storage space per linked list node # Simpler implementation Advantages of double linked list # Decrease in work when accessing a random node # Decrease in work when inserting or deleting a node
IN a circular doubly linked list with 10 nodes what is the minimum number of links that might have to be changed if we want to delete a node other than the head node?
Two, but there is no "might" about it. For any node (including the head) we must update exactly two links: node.next.prev = node.prev node.prev.next = node.next If the node is also the head then we also update the head node pointer: head = node.next Aside from deleting the head node, the only special case we need to deal with is when the node being deleted is the only node in the list because this will refer to itself (thus head = node.next would result in head pointing to a node we're about to delete). However, we can avoid this special case by using a sentinel node to represent the head of the list. In this way, every list, including an empty list, always has at least one node; the sentinel. Note that with a circular doubly linked list there no need to keep track of the tail node given that the head node's previous node already refers to the tail. However, with a singly-linked circular list we only need to keep track of the tail node since the next node after the tail is always the head node.
Algoritm for deleting the last element from a list?
Given a list and a node to delete, use the following algorithm: // Are we deleting the head node? if (node == list.head) { // Yes -- assign its next node as the new head list.head = node.next } else // The node is not the head node { // Point to the head node prev = list.head // Traverse the list to locate the node that comes immediately before the one we want to delete while (prev.next != node) { prev = prev.next; } end while // Assign the node's next node to the previous node's next node prev.next = node.next; } end if // Before deleting the node, reset its next node node.next = null; // Now delete the node. delete node;
Is deleting the only node in linked list a special case?
It depends on whether or not you have a head node. If you do, then it is not a special case. If you don't, then it is, and you need to be able to update the head pointer.
How can you close the iframe in calling page in php and javascript?
Using JavaScript, you can accomplish this by deleting the iframe node from the DOM.
What is time complexity to delete a node in singly linked list?
Time complexity to locate a node in a list of n nodes is O(1) at best (first node), O(n) at worst (last node) and O(n/2) on average. Once located, deleting a node takes constant time O(1).
How do you delete node from acircular linked list?
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).
Explain any two advantages using Single linked list over Doubly linked list and vice-versa?
Advantages of single linked list: # Decrease in storage space per linked list node # Simpler implementation Advantages of double linked list # Decrease in work when accessing a random node # Decrease in work when inserting or deleting a node
IN a circular doubly linked list with 10 nodes what is the minimum number of links that might have to be changed if we want to delete a node other than the head node?
Two, but there is no "might" about it. For any node (including the head) we must update exactly two links: node.next.prev = node.prev node.prev.next = node.next If the node is also the head then we also update the head node pointer: head = node.next Aside from deleting the head node, the only special case we need to deal with is when the node being deleted is the only node in the list because this will refer to itself (thus head = node.next would result in head pointing to a node we're about to delete). However, we can avoid this special case by using a sentinel node to represent the head of the list. In this way, every list, including an empty list, always has at least one node; the sentinel. Note that with a circular doubly linked list there no need to keep track of the tail node given that the head node's previous node already refers to the tail. However, with a singly-linked circular list we only need to keep track of the tail node since the next node after the tail is always the head node.
When drawing a Data Flow Diagram How do you represent deleting data?
A DFD consists of a set of processes and data stores and shows the flow of data through the system. First create a process node on the diagram and label it that the 'process' is to delete data. If the data is being deleted from a data store then draw an arrow to the data store that the information is being deleted from and label the arrow with the data being deleted. If the data is being deleted from the data flowing into the process node then show the data before the node and then remove the data after the node.
What is a negative application condition?
Negative Application Condition is a term widely used in Graph Transformation System. Informally, transition from one graph to other (such as deleting/inserting a node/edge) occurs only when this condition is not true.
Write an iterative function to search an element in a binary search tree?
_node* search (_node* head, _key key) { _node* node; for (node=head; node != NULL;;) { if (key == node->key) return node; else if (key < node.>key) node = node->left; else node = node->right; } return node; }
How to Print data of all nodes of linked list?
for (node=head; node!=null; node=node->next) printnode(node);
How to find the mirror image of a binary tree?
Refer to http://cslibrary.stanford.edu/110/BinaryTrees.html void mirror(struct node* node) { if (node==NULL) { return; } else { struct node* temp; // do the subtrees mirror(node->left); mirror(node->right); // swap the pointers in this node temp = node->left; node->left = node->right; node->right = temp; } }
