SingleLinkedList head D Node next - data- String value = "Tom" Node next- data- String value="Dick" Node nodeRef = nodeRef.next; nodeRef.next = new Node("Peter"); next- data- String value = "Harry" What change will be made to the list after the following codes are run? Node nodeRef = head; While (nodeRef.next!=null) Node next null data- String value="Sam"

SingleLinkedList head D Node next - data- String value = "Tom" Node next- data- String value="Dick" Node nodeRef = nodeRef.next; nodeRef.next = new Node("Peter"); next- data- String value = "Harry" What change will be made to the list after the following codes are run? Node nodeRef = head; While (nodeRef.next!=null) Node next null data- String value="Sam"

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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get_points("safe", "loft") // => List(A,A,C,A)get_points("safe", "gate") // => List(A,C,A,C)get_points("safe", "star") // => List(C,A,P,A)get_points("safe", "sums") // => List(C,A,A,P) Create a function called get_points which calculates the returns ,as shown above in the tests, by comparing the second word with the first word. The get_points function should output a list of 4 elements of type Letter.if a letter is present in the first word and in the correct place it should return with a "C". If the letter is present but not in the right place it should be a "P".And if the letter is not in the hidden_word then it should an "A" To do this you need to use the finder function to calculate if all the letters that are present in the first wordex. finder("safe", "loft") -> List(f) using the fix_spot function recursively analyse whether if the letter should be A,C or P. The finder function is a wrapper for the fix_spot function calling the fix_spot function with appropriate…
card_t * moveCardBack (card t *head); The moveCardBack function will take the card in front of the pile and place it in the back. In coding terms, you are taking the head of the linked list and moving it to the end. The function has one parameter which is the head of the linked list. After moving the card to the back, the function returns the new head of the linked list.
Below you're given a Node class and a LinkedList class. You will implement a method for the LinkedList class named "delete48in148". That is, whenever there is a sequence of nodes with values 1, 4, 8, we delete the 4 and 8. For exCample, Before: 1 -> 4 -> 8 LAfter: 1 Before: 7 -> 1 -> 4 -> 8 -> 9 -> 4 -> 8 After: 7 -> 1 -> 9 -> 4 -> 8 Before: 7 -> 1 -> 4 -> 8 -> 4 -> 8 -> 4 -> 8 -> 9 After: 7 -> 1 -> 9 Note from the above example that, after deleting one instance of 48, there may be new instances of 148 formed. You must delete ALL of them. Requirement: Your implementation must be ITERATIVE (i.e., using a loop). You must NOT use recursion. Recursive solutions will NOT be given marks. import ... # No other import is allowed
using namespace std; class SinglyLinkedListNode { // INSERT YOUR CODE HERE }; class SinglyLinkedList { public: SinglyLinkedListNode *head; SinglyLinkedListNode *tail; SinglyLinkedList() { this->head = nullptr; this->tail = nullptr; } voidinsert_node(intnode_data) { // INSERT YOUR CODE HERE } }; void free_singly_linked_list(SinglyLinkedListNode* node) { // INSERT YOUR CODE HERE } // Complete the has_cycle function below. /* * For your reference: * * SinglyLinkedListNode { * int data; * SinglyLinkedListNode* next; * }; * */ bool has_cycle(SinglyLinkedListNode* head) { SinglyLinkedListNode* temp = head; bool isCycle = false; while (temp != nullptr) { // INSERT YOUR CODE HERE } } int main() { // INSERT YOUR CODE HERE TO TEST YOUR CODE return0; }
The below code declares a new Vector named 'vec'. What is the size of the vector at the end of the execution? Vector vec = new Vector(); for(int i=1;i<=10; i++) { vec.addElement(i); } vec.removeElement(9); vec.removeElement(10); vec.removeElement(11);
Assign negativeCntr with the number of negative values in the linked list. Thanks. // ===== Code from file IntNode.java =====public class IntNode {private int dataVal;private IntNode nextNodePtr; public IntNode(int dataInit, IntNode nextLoc) {this.dataVal = dataInit;this.nextNodePtr = nextLoc;} public IntNode(int dataInit) {this.dataVal = dataInit;this.nextNodePtr = null;} /* Insert node after this node.* Before: this -- next* After: this -- node -- next*/public void insertAfter(IntNode nodePtr) {IntNode tmpNext; tmpNext = this.nextNodePtr; // Remember nextthis.nextNodePtr = nodePtr; // this -- node -- ?nodePtr.nextNodePtr = tmpNext; // this -- node -- next} // Grab location pointed by nextNodePtrpublic IntNode getNext() {return this.nextNodePtr;}public int getDataVal() {return this.dataVal;}}// ===== end ===== // ===== Code from file CustomLinkedList.java =====import java.util.Random; public class CustomLinkedList {public static void main(String[] args) {Random randGen = new…
Create a queue, size of queue will be dependent on the user. Insert the numbers in the queue till the queue reaches the size. Create a menu and perform the following function on that queue.Enqueue: Add an element to the end of the queueDequeue: Remove an element from the front of the queueIsEmpty: Check if the queue is emptyIsFull: Check if the queue is fullPeek: Get the value of the front of the queue without removing it
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Missing Brackets Checking: Write a technique to find on the off chance that an articulation has missing sections or not utilizing StackLinkedList. The technique takes a String articulation as a boundary and return True on the off chance that the articulation's sections are right and False
KWLinkedList<Integer> list = new KWLinkedList<Integer> (); for(int i = 1; i <= 4; i++) list.add( i ); ListIterator <Integer> iter = list.listIterator(); while(iter.hasNext()) { int x = iter.next(); int y = iter.next(); iter.set(x + y); } while(iter.hasPrevious()) System.out.print(iter.previous() + " ");
Consider the following linked list of the form. This list already exists, so you do not need to write the code to create the list below: struct nodeType { int infoData; nodeType * next; }; nodeType *first; … and containing the values (see image) : Write a code segment that prints all the current values in the linked list.
newNodePtr = malloc (sizeof (MenuNode) ) newNodePtr->data = inputData; newNodeFtr->next = NULL; if (front==NULL) front = newNodePtr else newNodePtr->next = front; front = newNodePtr; return 0, Figure Q2 (a) Write a program segment to display the content of the linked list.