Using Java programming language solve the following: SLL CLASS: //************************** SLL.java ********************************* // a generic singly linked list class public class SLL { private class SLLNode { private T info; private SLLNode next; public SLLNode() { this(null,null); } public SLLNode(T el) { this(el,null); } public SLLNode(T el, SLLNode ptr) { info = el; next = ptr; } } protected SLLNode head, tail; public SLL() { head = tail = null; } public boolean isEmpty() { return head == null; } public void addToHead(T el) { head = new SLLNode(el,head); if (tail == null) tail = head; } public void addToTail(T el) { if (!isEmpty()) { tail.next = new SLLNode(el); tail = tail.next; } else head = tail = new SLLNode(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 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 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 tmp = head; while(tmp != null){ str += tmp.info + " "; tmp = tmp.next; } return str+"]"; } public boolean contains(T el) { if(head == null) return false; SLLNode 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 p = head; while(p != null) { count++; p = p.next; } return count; } } Part I: Programming Use the singly linked list class (Given below) to implement integers of unlimited size. Each node of the list should store one digit of the integer. You are required to implement the addition, subtraction and multiplication operations. In addition, provide a test class that will do the following: It will keep asking the user to enter two integers, choose one of the operations (addition, subtraction, multiplication) and then display the result. The program will stop upon e.

1. Using Java programming language solve the following:
SLL CLASS:
//************************** SLL.java *********************************
// a generic singly linked list class
public class SLL {
private class SLLNode {
private T info;
private SLLNode next;
public SLLNode() {
this(null,null);
}
public SLLNode(T el) {
this(el,null);
}
public SLLNode(T el, SLLNode ptr) {
info = el; next = ptr;
}
}
protected SLLNode head, tail;
public SLL() {
head = tail = null;
}
public boolean isEmpty() {
return head == null;
}
public void addToHead(T el) {
head = new SLLNode(el,head);
if (tail == null)
tail = head;

2. }
public void addToTail(T el) {
if (!isEmpty()) {
tail.next = new SLLNode(el);
tail = tail.next;
}
else head = tail = new SLLNode(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 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;

3. head = head.next; // and el is in the head node;
else { // if more than one node in the list
SLLNode 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 tmp = head;
while(tmp != null){
str += tmp.info + " ";
tmp = tmp.next;
}
return str+"]";
}
public boolean contains(T el) {
if(head == null)
return false;
SLLNode tmp = head;
while(tmp != null){
if(tmp.info.equals(el))
return true;

4. tmp = tmp.next;
}
return false;
}
public int size(){
if(head == null)
return 0;
int count = 0;
SLLNode p = head;
while(p != null) {
count++;
p = p.next;
}
return count;
}
}
Part I: Programming Use the singly linked list class (Given below) to implement integers of
unlimited size. Each node of the list should store one digit of the integer. You are required to
implement the addition, subtraction and multiplication operations. In addition, provide a test
class that will do the following: It will keep asking the user to enter two integers, choose one of
the operations (addition, subtraction, multiplication) and then display the result. The program
will stop upon entering the "#" character (without the double quotes). Part II: Problem Solving
With respect to the programming problem in Part I , What is the tightest asymptotic running time
for each of your operations, expressed in terms of the number of digits for the two operands, m
and n, of each operation? Clearly justify your answer.