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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Methods of StringBuilder class

The StringBuilder class can be used to create mutable strings. The methods defined in the StringBuilder class are widely used to manipulate and concatenate strings. It is a non-synchronized class which has public methods that can be accessed by multiple threads. It is not thread-safe.

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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…
] ] 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); }