Given the following partial code, fill in the blank to complete the code necessary to remove last node. (don't forget the semicolon) class Node { public Object data = null; public Node next = null; } Node head = new Node(); // first Node head.next = new Node(); // second Node head.next.next = new Node(); // third node head.next.next.next = new Node(); // fourth node //

Given the following partial code, fill in the blank to complete the code necessary to remove last node. (don't forget the semicolon) class Node { public Object data = null; public Node next = null; } Node head = new Node(); // first Node head.next = new Node(); // second Node head.next.next = new Node(); // third node head.next.next.next = new Node(); // fourth node //

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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For any element in keysList with a value greater than 50, print the corresponding value in itemsList, followed by a comma (no spaces). Ex: If the input is: 32 105 101 35 10 20 30 40 the output is: 20,30, 1 #include 2 3 int main(void) { const int SIZE_LIST = 4; int keysList[SIZE_LIST]; int itemsList[SIZE_LIST]; int i; 4 6 7 8 scanf("%d", &keysList[0]); scanf ("%d", &keysList[1]); scanf("%d", &keysList[2]); scanf("%d", &keysList[3]); 10 11 12 13 scanf ("%d", &itemsList[0]); scanf ("%d", &itemsList[1]); scanf("%d", &itemsList[2]); scanf ("%d", &itemsList[3]); 14 15 16 17 18 19 /* Your code goes here */ 20 21 printf("\n"); 22 23 return 0; 24 }
The numbers on the left margin denote line numbers. line#01 class LL_node {02 int val;03 LL_node next;04 public LL_node (int n) {05 val = n;06 next = null;07 }08 public void set_next (LL_node nextNode) {09 next = nextNode;10 }11 public LL_node get_next () {12 return next;13 }14 public void set_value (int input) {15 val = input;16 }17 public int get_value () {18 return val;19 }20 }21 public class LL {22 protected LL_node head;23 protected LL_node tail;24 public LL () {25 head = null;26 tail = null;27 }28 public int append (int n) {29 if (head == null) {30 head = new LL_node(n);31 tail = head;32 } else {33 LL_node new_node;34 new_node = new LL_node(n);35 tail.set_next (new_node);36 tail = new_node;37 }38 return n;39 }40 } which the following statement…
//Can you explain how this code work or the meaning of this code?#include <iostream>using namespace std;struct Node {int data;struct Node *next;};struct Node* head = NULL;struct Node* rhead = NULL;int count1;//insert at front for reverselistvoid insertfront(int data){Node* new_node = (Node*) malloc(sizeof(Node)); new_node->data = data; new_node->next = rhead; rhead=new_node;}//insert at endvoid insert(int new_data){Node* ptr;ptr = head;Node* new_node = (Node*) malloc(sizeof(Node)); new_node->data = new_data; new_node->next = NULL;if (head == NULL) { head = new_node; } else{while (ptr->next!= NULL) { ptr = ptr->next;}ptr->next=new_node;}count1++;}//display listvoid display() {struct Node* ptr;ptr = head;if(head==NULL){cout<<"Sorry the list is empty";return;}while (ptr != NULL) {cout<< ptr->data <<" ";ptr = ptr->next;}cout<<endl;}//delete datavoid deletedata(int data){Node* ptr,*ptrpre;ptr = head;if (head == NULL) {…
] ] get_nhbr In the cell below, you are to write a function called "get_nhbr(Ist, graph)" that takes in two inputs: a list of vertices and a graph. The function is to return a list that contains the neighborhood of the vertices in 'Ist' (remember that this means you are finding the union of the individual vertices' neighborhoods). + Code + Markdown After compiling the above cell, you should be able to compile the following cell and obtain the desired outputs. print (get_nhbr(["A", "D"], {"A" : ["B"], "B" : ["A", "D", "E"], "C" : ["E"], "D":["B"], "E":["B","C","F"], "F":["E"]}), get_nhbr(["B", "C", "F"], {"A" : ["B"], "B" : ["A", "D", "E"], "C" : ["E"], "D":["B"], "E":["B","C","F"], "F":["E"]})) This should return ["B"] ["A", "D", "E"] Python Python
#include using namespace std; struct ListNode { string data; ListNode *next; }; int main() { ListNode *p, *list; list = new ListNode; list->data = "New York"; p new ListNode; p->data = "Boston"; list->next = p; p->next = new ListNode; p->next->data = "Houston"; p->next->next = nullptr; // new code goes here Which of the following code correctly deletes the node with value "Boston" from the list when added at point of insertion indicated above? O list->next = p; delete p; O p = list->next; %3D list->next = p->next; delete p; p = list->next; list = p->next; delete p; O None of these O p = list->next; %3D list->next = p; %3D delete p;
Explain the following code : public static void printLeafNodes(TreeNode node) { // base case if (node == null) { return; } if (node.left == null && node.right == null) { System.out.printf("%d ", node.value); } printLeafNodes(node.left); printLeafNodes(node.right); }
C++ PROGRAMMINGTopic: Binary Search Trees Explain the c++ code below.: It doesn't have to be long, as long as you explain what the important parts of the code do. (The code is already implemented and correct, only the explanation needed) node* left(node* p) { return p->left; } node* right(node* p) { return p->right; } node* sibling(node* p){ if(p != root){ node* P = p->parent; if(left(P) != NULL && right(P) != NULL){ if(left(P) == p){ return right(P); } return left(P); } } return NULL; } node* addRoot(int e) { if(size != 0){ cout<<"Error"<<endl; return NULL; } root = create_node(e,NULL); size++; return root; } node* addLeft(node* p, int e) { if(p->left == NULL){ node* newLeft =…