Which of the following is true about linked lists with header and trailer nodes? a) Using header and trailer nodes simplifies the implementation because it eliminates the special cases. b) None of these c) Using header and trailer nodes saves some important information in the header and trailer nodes. d) The header (or trailer) node is the same thing as a head (or tail) reference. e) Using header and trailer nodes can reduce the time complexity of the code.
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- TRUE or FALSE? Answer the following question and state the reason why: The delete operation only involves the removing of the node from the list without breaking the links created by the next node. You need an array to represent each node in a linked list. STL lists are also efficient at adding elements at their back because they have a built-in pointer to the last element in the list. A circular linked list has 2 node pointers. cout<<list.back()<<endl; = The back member function returns a reference to the last element in the list. In a Dynamic Stack, the pointer top stays at the head after a push operation. During a Pop operation in Static Stack, the elements are being moved one step up. In a dynamic implementation of stack, the pointer top has an initial value of null. In a dynamic stack, the node that was popped is deleted. In a dynamic stack, the pointer top stays at the head after push operation. STL function top returns a reference to element at the top of the…Computer Science Describe a situation where you may wish to create a link list of ITERATORS. Is it possible to create a link list of various types of iterators? For example a list with list iterators, map iterators, and set iterators?Provide full C++ Code: Your code must have the following three files: Playlist.hpp - Class declaration for a linked list of Nodes (very similar to lab 08 but with some changes to methods, as described below) with following private data members: private:Node* first;Node* last;int size; Playlist.cpp - Class definition main.cpp - main() function Checkpoint A Build a playlist (of songs) by reading data members of several songs from a file and printing them. Implement the following methods: PlayList(); //Default constructorPlayList(string filename); //Parameterized constructor that reads data members of song from a file and builds the PlayList~PlayList(); //Deletes all nodes of the linked listbool InsertNodeLast(Node *myNewNode); //Inserts myNewNode at the end of the linked listbool InsertNodeFirst(Node *myNewNode); //Inserts myNewNode at the start of the linked listbool DeleteFirst(); //Deletes the first node of the linked listint Size() const { return size; }friend ostream&…
- Can someone help me with this? C++ programming please! Given the MileageTrackerNode class, complete main() to insert nodes into a linked list (using the InsertAfter() function). The first user-input value is the number of nodes in the linked list. Use the PrintNodeData() function to print the entire linked list. DO NOT print the dummy head node. Ex. If the input is: 3 2.2 7/2/18 3.2 7/7/18 4.5 7/16/18 the output is: 2.2, 7/2/18 3.2, 7/7/18 4.5, 7/16/18 Main.cpp #include "MileageTrackerNode.h"#include <string>#include <iostream>using namespace std; int main (int argc, char* argv[]) {// References for MileageTrackerNode objectsMileageTrackerNode* headNode;MileageTrackerNode* currNode;MileageTrackerNode* lastNode; double miles;string date;int i; // Front of nodes listheadNode = new MileageTrackerNode();lastNode = headNode; // TODO: Read in the number of nodes // TODO: For the read in number of nodes, read// in data and insert into the linked list // TODO: Call the…Write a program which will concatenate two single linked lists. The structure is defined below. Assume that head1 contains the starting address of the first list and head2 contains the starting address of the second list. You need to join them together so that all the students’ roll no and names of both the lists will be displayed together with the help of head1 pointer. struct link { int rollno; char name[20]; struct link *next; }; typedef struct link node; node *head1=NULL, *head2=NULL; //global declarationANSWER THAT MCQ QUESTION IN 2 MINUTES PLEASE.. The following function reverse() is supposed to reverse a singly linked list. There is one line missing at the end of the function. /* Link list node */ struct node { int data; struct node* next; }; /* head_ref is a double pointer which points to head (or start) pointer of linked list */ static void reverse(struct node** head_ref) { struct node* prev = NULL; struct node* current = *head_ref; struct node* next; while (current != NULL) { next = current->next; current->next = prev; prev = current; current = next; } /*ADD A STATEMENT HERE*/ }
- Write a function that changes the places of the first and last nodes of a linked list. I.e. First node will become last node, and the last node will become first node. Others nodes will not be affected.THE FUNCTION WILL NOT EXCHANGE DATA, IT WILL CHANGE LINKS.What description to Array and Linked List is mistake ? a. Using Linked list, if there is a existed data will be removed, it needs only to move the links of data node and doesn't need to change the location of data saved. b. Using Linked list, if there is a new data will be increased, it needs only to move the links of data node and doesn't need to change the location of data saved. c. If the size of main memory is enough, there is no upper bound limitation to increase new data when Linked list is used. d. For limited number of data, Array is allocated less memory size, because all sequential data are allocated into continuously space e. Using Array, if there is a new data will be increased, it needs to move some existed data to new location to free the suitable location new data because the sequential order should be kept. f. If the size of main memory is enough, there is no upper bound limitation to increase new data when Array is used. g. The operations on Array is…Write a program in c++ which will concatenate two single linked lists. The structure is defined below. Assume that head1 contains the starting address of the first list and head2contains the starting address of the second list. You need to join them together so that allthe students’ roll number and name of both the lists will display together with the help of head1pointer.struct link{int rollno;char name[20];struct link *next;};typedef struct link node;node *head1=NULL, *head2=NULL; //global declaration
- (IntelliJ) Write an easier version of a linked list with only a couple of the normal linked list functions and the ability to generate and utilize a list of ints. The data type of the connection to the following node can be just Node, and the data element can be just an int. You will need a reference (Java) or either a reference or a pointer (C++) to the first node, as well as one to the last node. * Create a method or function that accepts an integer, constructs a node with that integer as its data value, and then includes the node to the end of the list. If the new node is the first one, this function will also need to update the reference to the first node. This function will need to update the reference to the final node. Consider how to insert the new node following the previous last node, and keep in mind that the next reference for the list's last node should be null. * Create a different method or function that iteratively explores the list, printing the int data values as…1-Let the list have a head and a tail. That is, a pointer (have a marker) to both the beginning (first Node) of the list and the last Node. What process does Tail facilitate? 2-insert(int index, int element): adds this element to the index position. For example, if index is 4, it adds this element between index 3 and 4 in the list. The size of the list has increased by one. 3-append(int elem): Adds the element to the end of the list. The size of the list has increased by one. 4-get(int index): Returns the element at the index position of the list, no change in the list. 5-remove(int index): Returns the element at the index position of the list. This element is removed from the list and the list size is reduced by one. 6-findMin(): returns the index of the smallest number in the list. 7-findMax(): returns the index of the largest number in the list. 8-search(int elem): searches elem in the list. It returns -1 when you can't find elem's index when you find it. 9-ToArray(): Return an…1-Let the list have a head and a tail. That is, a pointer (have a marker) to both the beginning (first Node) of the list and the last Node. What process does Tail facilitate? 2-insert(int index, int element): adds this element to the index position. For example, if index is 4, it adds this element between index 3 and 4 in the list. The size of the list has increased by one. 3-append(int elem): Adds the element to the end of the list. The size of the list has increased by one. 4-get(int index): Returns the element at the index position of the list, no change in the list. 5-remove(int index): Returns the element at the index position of the list. This element is removed from the list and the list size is reduced by one. 6-findMin(): returns the index of the smallest number in the list. 7-findMax(): returns the index of the largest number in the list. 8-search(int elem): searches elem in the list. It returns -1 when you can't find elem's index when you find it. 9-ToArray(): Return an…