Solve using Java programming language.
//************************** SLL.java *********************************
// a generic singly linked list class
public class SLL<T> {
private class SLLNode<T> {
private T info;
private SLLNode<T> next;
public SLLNode() {
this(null,null);
}
public SLLNode(T el) {
this(el,null);
}
public SLLNode(T el, SLLNode<T> ptr) {
info = el; next = ptr;
}
}
protected SLLNode<T> head, tail;
public SLL() {
head = tail = null;
}
public boolean isEmpty() {
return head == null;
}
public void addToHead(T el) {
head = new SLLNode<T>(el,head);
if (tail == null)
tail = head;
}
public void addToTail(T el) {
if (!isEmpty()) {
tail.next = new SLLNode<T>(el);
tail = tail.next;
}
else head = tail = new SLLNode<T>(el);
}
public T deleteFromHead() { // delete the head and return its info;
if (isEmpty())
return null;
T el = head.info;
if (head == tail) // if only one node on the list;
head = tail = null;
else head = head.next;
return el;
}
public T deleteFromTail() { // delete the tail and return its info;
if (isEmpty())
return null;
T el = tail.info;
if (head == tail) // if only one node in the list;
head = tail = null;
else { // if more than one node in the list,
SLLNode<T> tmp; // find the predecessor of tail;
for (tmp = head; tmp.next != tail; tmp = tmp.next);
tail = tmp; // the predecessor of tail becomes tail;
tail.next = null;
}
return el;
}
public void delete(T el) { // delete the node with an element el;
if (!isEmpty())
if (head == tail && el.equals(head.info)) // if only one
head = tail = null; // node on the list;
else if (el.equals(head.info)) // if more than one node on the list;
head = head.next; // and el is in the head node;
else { // if more than one node in the list
SLLNode<T> pred, tmp;// and el is in a nonhead node;
for (pred = head, tmp = head.next;
tmp != null && !tmp.info.equals(el);
pred = pred.next, tmp = tmp.next);
if (tmp != null) { // if el was found;
pred.next = tmp.next;
if (tmp == tail) // if el is in the last node;
tail = pred;
}
}
}
@Override
public String toString() {
if(head == null)
return "[ ]";
String str = "[ ";
SLLNode<T> tmp = head;
while(tmp != null){
str += tmp.info + " ";
tmp = tmp.next;
}
return str+"]";
}
public boolean contains(T el) {
if(head == null)
return false;
SLLNode<T> tmp = head;
while(tmp != null){
if(tmp.info.equals(el))
return true;
tmp = tmp.next;
}
return false;
}
public int size(){
if(head == null)
return 0;
int count = 0;
SLLNode<T> p = head;
while(p != null) {
count++;
p = p.next;
}
return count;
}
}
Given the code above, solve the following tasks (please show the solution for every task
clearly):
For part (c) this is the code for the test program:
public class SLL_Driver {
public static void main(String[] args) {
SLL<String> myList = new SLL<String>();
String[] cityNames = {"Jubail", "Riyadh", "Abha", "Dammam", "Taif"};
for(int i = 0; i < cityNames.length; i++)
myList.addToHead(cityNames[i]);
System.out.println("The list is: " + myList);
System.out.println("It is " + myLis.
Solve using Java programming language- ----------------------------.pdf
1. Solve using Java programming language.
//************************** SLL.java *********************************
// a generic singly linked list class
public class SLL<T> {
private class SLLNode<T> {
private T info;
private SLLNode<T> next;
public SLLNode() {
this(null,null);
}
public SLLNode(T el) {
this(el,null);
}
public SLLNode(T el, SLLNode<T> ptr) {
info = el; next = ptr;
}
}
protected SLLNode<T> head, tail;
public SLL() {
head = tail = null;
}
public boolean isEmpty() {
return head == null;
2. }
public void addToHead(T el) {
head = new SLLNode<T>(el,head);
if (tail == null)
tail = head;
}
public void addToTail(T el) {
if (!isEmpty()) {
tail.next = new SLLNode<T>(el);
tail = tail.next;
}
else head = tail = new SLLNode<T>(el);
}
public T deleteFromHead() { // delete the head and return its info;
if (isEmpty())
return null;
T el = head.info;
if (head == tail) // if only one node on the list;
head = tail = null;
else head = head.next;
return el;
}
public T deleteFromTail() { // delete the tail and return its info;
3. if (isEmpty())
return null;
T el = tail.info;
if (head == tail) // if only one node in the list;
head = tail = null;
else { // if more than one node in the list,
SLLNode<T> tmp; // find the predecessor of tail;
for (tmp = head; tmp.next != tail; tmp = tmp.next);
tail = tmp; // the predecessor of tail becomes tail;
tail.next = null;
}
return el;
}
public void delete(T el) { // delete the node with an element el;
if (!isEmpty())
if (head == tail && el.equals(head.info)) // if only one
head = tail = null; // node on the list;
else if (el.equals(head.info)) // if more than one node on the list;
head = head.next; // and el is in the head node;
else { // if more than one node in the list
SLLNode<T> pred, tmp;// and el is in a nonhead node;
for (pred = head, tmp = head.next;
tmp != null && !tmp.info.equals(el);
4. pred = pred.next, tmp = tmp.next);
if (tmp != null) { // if el was found;
pred.next = tmp.next;
if (tmp == tail) // if el is in the last node;
tail = pred;
}
}
}
@Override
public String toString() {
if(head == null)
return "[ ]";
String str = "[ ";
SLLNode<T> tmp = head;
while(tmp != null){
str += tmp.info + " ";
tmp = tmp.next;
}
return str+"]";
}
public boolean contains(T el) {
if(head == null)
return false;
5. SLLNode<T> tmp = head;
while(tmp != null){
if(tmp.info.equals(el))
return true;
tmp = tmp.next;
}
return false;
}
public int size(){
if(head == null)
return 0;
int count = 0;
SLLNode<T> p = head;
while(p != null) {
count++;
p = p.next;
}
return count;
}
}
Given the code above, solve the following tasks (please show the solution for every task
clearly):
For part (c) this is the code for the test program:
public class SLL_Driver {
6. public static void main(String[] args) {
SLL<String> myList = new SLL<String>();
String[] cityNames = {"Jubail", "Riyadh", "Abha", "Dammam", "Taif"};
for(int i = 0; i < cityNames.length; i++)
myList.addToHead(cityNames[i]);
System.out.println("The list is: " + myList);
System.out.println("It is " + myList.contains("Dammam") + " that the list contains Dammam.");
System.out.println("It is " + myList.contains("Jeddah") + " that the list contains Jeddah.");
}
}
1. (a) Comment the iterative public boolean contains(Element e) method of the given SLL T
class then implement it as a recursive method. Use an appropriate helper method in your
solution. (b) Comment the iterative public String toString( ) method of the given SLL T class
then implement it as a recursive method. Use appropriate helper method in your solution. (c) Use
the given test program to test your recursive methods. Sample program run: The list is: [ Taif
Dammam Abha Riyadh Jubail ] It is true that the list contains Dammam. It is false that the list
contains Jeddah.
