Linked List head M tail 10 15 20 25 30 35 null tail->next = head->next create_node(M) M->value = 40 M->next = head->next->next->next head->next->next = M M = head For x = 1 to 10 do Begin If (M->value % 2 = 0) then Output(M->value) M->value = M->value * x M = M->next %3D End

Linked List head M tail 10 15 20 25 30 35 null tail->next = head->next create_node(M) M->value = 40 M->next = head->next->next->next head->next->next = M M = head For x = 1 to 10 do Begin If (M->value % 2 = 0) then Output(M->value) M->value = M->value * x M = M->next %3D End

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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//main.cpp #include "linkedListType.h"int main(){Node* head = NULL;orderedLinkedList oll = orderedLinkedList(head); oll.insert(40);oll.insert(100);oll.insert(50); oll.printList(); oll.insert(50);oll.printList(); return 0;} //linkedListType.h #include <stdio.h>#include <iostream>using namespace std; class Node{public:int data;Node* next;}; class orderedLinkedList {Node* head; public:orderedLinkedList(Node* h) {head = h;} void insert(int new_data){Node* head_ref = head; while (head_ref != NULL && (head_ref->next != NULL) && (head_ref->next)->data < new_data){cout << head_ref->data << "\n";head_ref = head_ref->next;} if (head_ref != NULL && (head_ref->next != NULL) && (head_ref->next)->data == new_data){cout << "ERROR: Item to be inserted is already in the list\n";return;} /* Insert new node */Node* new_node = new Node();new_node->data = new_data;if (head_ref != NULL) {new_node->next =…
The programming language is Python Instructions: Create a function that will delete a node in a Linked List based on position number. On below example, if you want to delete position #2, it will remove the Banana (arrangement of nodes below is Apple, Banana, Cherry, Grapes, Orange). myLinkedList = LinkedList() myLinkedList.append("Banana") myLinkedList.append("Cherry") myLinkedList.append("Grapes") myLinkedList.append("Orange") myLinkedList.prepend("Apple") myLinkedList.deleteByPositionNum(2) node = myLinkedList.head while node: print(node.value, " ") node = node.next_node You may start with the function head:def deleteByPositionNum(self, positionNum):
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Given a linked list of N nodes such that it may contain a loop. A loop here means that the last node of the link list is connected to the node at position X(1-based index). If the link list does not have any loop, X=0. Remove the loop from the linked list, if it is present, i.e. unlink the last node which is forming the loop. Example 1: Input: N = 3 value[] = {1,3,4} X = 2 Output: 1 Explanation: The link li. st looks like 1 -> 3 -> 4 ^ | |____|
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package chapter02;  public class LinkedList {  protected LLNode list;   public LinkedList()  {  list = null;  }   public void addFirst(T info)  {  LLNode node = new LLNode(info);  node.setLink(list);  list = node;  }   public void addLast(T info)  {  LLNode curr = list;   LLNode newNode = new LLNode(info);  if(curr == null)  {  list = newNode;  }  else  {  while(curr.getLink() != null)  {  curr = curr.getLink();  }  curr.setLink(newNode);  }  }   public void add(T afterThis, T info)  {  //TODO Complete this method as required in the homework instructions    }   public void removeFirst()  {  //TODO Complete this method as required in the homework instructions    }  public void removeLast()  {  //TODO Complete this method as required in the homework instructions    }   public int size()  {  //TODO Complete this method…
class Node { public: Node() : data(0), prev(nullptr), next(nullptr) {} Node(int theData, Node* prevLink, Node* nextLink) : data(theData), prev(prevLink), next(nextLink) {} int getData() const { return data; } Node* getPrev() const { return prev; } Node* getNext() const { return next; } void setData(int theData) { data = theData; } void setPrev(Node* prevLink) { prev = prevLink; } void setNext(Node* nextLink) { next = nextLink; } ~Node(){} private: int data; Node* prev; Node* next; }; class AnyList { // friend function overloads the insertion operation public: // copy constructor // overloaded assignment operator // other member functions not necessary for your implementation private: Node *first; Node *last; int count; }; please help write copy constructor #include "AnyList.h" #include <iostream> using namespace std; // Definition of copy constructor // Instructions omitted intentionally. // Your code
Which code part (A, B, C, or D) is incorrect? Add element in Linked List at particular index: add(int index, T item) void add(int index, int value) { // A. ============================================= if(index < 0 || index > count){ cerr << "Index Out of Bounds"; return; } node *temp = new node; temp->item = value; // A. ============================================= // B. ============================================= if (index == 0) { temp->next = head; head = temp; } // B. ============================================= // C. ============================================= else { node *previous = head; for(int i = 0; i < index - 1; ++i){ previous = previous->next; } // C. ============================================= // D. ============================================= previous->next = temp; temp->next = previous->next; } // D. ============================================= ++count; return; }
#ifndef NODES_LLOLL_H#define NODES_LLOLL_H #include <iostream> // for ostream namespace CS3358_SP2023_A5P2{ // child node struct CNode { int data; CNode* link; }; // parent node struct PNode { CNode* data; PNode* link; }; // toolkit functions for LLoLL based on above node definitions void Destroy_cList(CNode*& cListHead); void Destroy_pList(PNode*& pListHead); void ShowAll_DF(PNode* pListHead, std::ostream& outs); void ShowAll_BF(PNode* pListHead, std::ostream& outs);} #endif #include "nodes_LLoLL.h"#include "cnPtrQueue.h"#include <iostream>using namespace std; namespace CS3358_SP2023_A5P2{ // do breadth-first (level) traversal and print data void ShowAll_BF(PNode* pListHead, ostream& outs) { cnPtrQueue queue; CNode* currentNode; queue.push(pListHead->data); while (!queue.empty()) { currentNode = queue.front(); queue.pop(); if…
#ifndef NODES_LLOLL_H#define NODES_LLOLL_H #include <iostream> // for ostream namespace CS3358_SP2023_A5P2{ // child node struct CNode { int data; CNode* link; }; // parent node struct PNode { CNode* data; PNode* link; }; // toolkit functions for LLoLL based on above node definitions void Destroy_cList(CNode*& cListHead); void Destroy_pList(PNode*& pListHead); void ShowAll_DF(PNode* pListHead, std::ostream& outs); void ShowAll_BF(PNode* pListHead, std::ostream& outs);} #endif #include "nodes_LLoLL.h"#include "cnPtrQueue.h"#include <iostream>using namespace std; namespace CS3358_SP2023_A5P2{ // do breadth-first (level) traversal and print data void ShowAll_BF(PNode* pListHead, ostream& outs) { cnPtrQueue queue; CNode* currentNode; queue.push(lloLLPtr->getHead()); while (!queue.empty()) { currentNode = queue.front(); queue.pop(); if…
Data and links are node class variables. Which boolean expression determines whether two nodes (p and q) have the same numbers? Let p and q be non-null. p.data() q.data() p.link-q.link No answers here
Write a code for a simple text editor/Nodepad. Use fully connected linked list, with four pointers next,prev,up and down. Note: Use linkedlist DSA C++ and also file handling. Please complete the requirements. I asked this question multiplie times before. I don't want incorrect answer this time.