How can I add it and delete Node First and delete Node Last in a double-link list
Just want to add delete Node First and delete Node Last. I did not know how to add it in your programming, but I want to help it solve it
i have 3 class
class node .
class doublyLinkedListMain .
class doublyLinkedList .
I want to add special programming in class doublyLinkedList this one delete Node First and delete Node Last
class doublyLinkedList
{
Node head;
public void push(int newdata)
{
Node NewNode = new Node(newdata);
NewNode.next = head;
NewNode.prev = null;
if (head != null) head.prev = NewNode;
head = NewNode;
}
public void insertAfter(Node PrevNode, int newdata)
{
if (PrevNode == null)
{
System.out.println("The given previous node cannot be null");
return;
}
Node NewNode = new Node(newdata);
NewNode.next = PrevNode.next;
PrevNode.next = NewNode;
NewNode.prev = PrevNode;
if (NewNode.next != null)
NewNode.next.prev = NewNode;
}
public void append(int newdata)
{
Node NewNode = new Node(newdata);
Node last = head;
NewNode.next = null;
if (head == null)
{
NewNode.prev = null;
head = NewNode;
return;
}
while (last.next != null) last = last.next;
last.next = NewNode;
NewNode.prev = last;
return;
}
void insertBefore(Node NextNode, int newdata)
{
if (NextNode == null)
{
System.out.println("the given next node cannot be NULL");
return;
}
Node NewNode = new Node(newdata);
NewNode.data = newdata;
NewNode.prev = NextNode.prev;
NextNode.prev = NewNode;
NewNode.next = NextNode;
if (NewNode.prev != null)
NewNode.prev.next = NewNode;
else head = NewNode;
}
void deleteNode( Node del)
{
if (head == null || del == null) return;
if (head == del) head = head.next;
if (del.next != null) del.next.prev = del.prev;
if (del.prev != null) del.prev.next = del.next;
return;
}
void printList()
{
Node n = head;
while (n != null)
{
System.out.print(n.data+" ");
n = n.next;
}
System.out.print(" \n ") ;
}
}
sorry, but you have different mistakes in your code. Firstly, for a list you need two or three pointers as attribute in your List class (Node head /* begin */, tail /* end */, current /* yes, the current element on which you have access */ - I have learned it with a current pointer but depending on your implementation you can do it without it too). Because of that, you must change some of your code. For example, you need a hasAccess() : boolean method, if you use a current pointer, which return current != null and your methods must consider and use tail. Furthermore, you should add a method isEmpty():
public boolean isEmpty() {
return head == null;
}
This method checks if the list is empty. In the case, that the list is empty (the method returns true), you cannot do some things, but you had seen that yet. But, because of the new pointer tail, you have its much easier to delete the last Node (see below).
Please, change all your attributes to private and code a getter and a setter for each of them. For example in the class Node for the attribute next:
private Node next;
// other code...
public Node getNext() {
return next;
}
public void setNext(Node next) {
this.next = next;
}
I have solved the problem with "delete Node First and delete Node Last" in the following way:
public class DoublyLinkedList {
private Node head; // begin
private Node tail; // end
public void deleteFirstElement() {
if (!isEmpty()) {
head = head.getNext();
}
}
public void deleteLastElement() {
if (!isEmpty()) {
tail = tail.getPrev();
}
}
// other code
}
Of course, you have to proof your other code, if it works. I hope that I could help you.
I have a LinkedList (Own code for the LinkedList) with char's in it. This is the complete list: ['a','b','I','d','R','A','7','p'].
I am trying to write a method which will deleting all characters that is NOT a upper case letter. After running the method, the LinkedList should look like this ['I','R','A'].
But after running my code I get the same list as return, this list: ['a','b','I','d','R','A','7','p'].
Here is my code for the method:
public static ListNode copyUpperCase(ListNode head) {
ListNode ptr = head;
while(!isEmpty(ptr.next)){
if(!Character.isUpperCase(ptr.element)){
ptr = ptr.next.next;
//System.out.println(ptr.element);
}
ptr = ptr.next;
}
return head;
}
Here is isEmpty():
public static boolean isEmpty(ListNode l) {
if ( l == null )
throw new ListsException("Lists: null passed to isEmpty");
return l.next == null;
}
Here is ListNode:
public class ListNode {
public char element;
public ListNode next;
}
I can see that the search part is working, but I can't get the deleting the node part right, any suggestions?
public static ListNode copyUpperCase(ListNode head) {
ListNode ptr = head;
while(!isEmpty(ptr.next)){
if(!Character.isUpperCase(ptr.element)){
ptr.next = ptr.next.next;
//System.out.println(ptr.element);
}
ptr = ptr.next;
}
return head;
}
You need to "CHANGE" the list, so you just missed the assignment to the element and not the the local variable
THIS code will not work however, as this just assign to the next element, without looking if next element actully is a good one, and then skips to that one
Edit: full working code
class ListNode {
public ListNode(char element,ListNode next ) {
this.element = element;
this.next = next;
}
public char element;
public ListNode next;
void print() {
System.out.print(this.element+",");
if(this.next != null) {
this.next.print();
}
else {
System.out.println("");
}
}
}
public class main {
//Imo you should only check if this elem is a null one, as a null means empty, a null on next only means that it's the last elem, but will still contain data
public static boolean isEmpty(ListNode l) {
return l == null;
}
public static ListNode getNextUpper(ListNode head) {
while(!isEmpty(head)){
if(Character.isUpperCase(head.element)) {
return head;
}
head = head.next;
}
return null;
}
public static ListNode copyUpperCase(ListNode head) {
ListNode newhead = getNextUpper(head);
ListNode temp = newhead;
while(!isEmpty(temp)){
temp.next = getNextUpper(temp.next);
temp = temp.next;
}
return newhead;
}
public static void main(String[] args) {
ListNode t = new ListNode('a' , new ListNode('b' , new ListNode('I', new ListNode('d', new ListNode('R', new ListNode('A', new ListNode('7', new ListNode('p',null))))))));
t.print();
ListNode newt = copyUpperCase(t);
newt.print();
}
}
ptr is a local variable, so
ptr = ptr.next.next;
doesn't modify your linked list.
You should modify ptr.next instead. Besides that, you may have to modify head, if the original head doesn't refer to an upper case letter.
Something like this should work :
// find the first valid (upper case) element and set head to refer to it
while (!Character.isUpperCase(head.element) && head != null)
head = head.next;
ListNode ptr = head;
if (ptr != null)
// eliminate all non-upper case elements
while(!isEmpty(ptr.next)){
if(!Character.isUpperCase(ptr.next.element)){
ptr.next = ptr.next.next;
}
ptr = ptr.next;
}
}
return head;
Data structures class, implementing a singly linked-list with head, tail and current nodes. Having trouble with a method, could use a nudge in the right direction.
From the assignment, write the method:
add( item ) : adds the item (String) after the current node in the list and sets the current pointer to refer to the new node.
My attempt:
public void add(String item)
{
if(curr != null)
{
Node newNode = new Node(item, curr.next);
curr.next = newNode;
curr = newNode;
}
else
{
head = tail = new Node(item, null);
curr = head;
}
}
My add method only seems to work when I'm adding items to the middle of the list, not on either end. If I use it to add a few items and then print the list, only the first one I added will be on the list, while my prepend and append methods have tested just fine.
Is there any glaring issue with my code? I feel like I'm missing something obvious.
All:
public class LinkedList {
Node head = null; /* Head of the list */
Node tail = null; /* Tail of the list */
Node curr = null; /* Current node in the list */
public void prepend(String item) {
if (head == null) {
head = tail = new Node(item, null);
curr = head;
} else {
head = new Node(item, head);
curr = head;
}
}
public void append(String item) {
if (head == null) {
head = tail = new Node(item, null);
curr = tail;
} else {
tail.next = new Node(item, null);
tail = tail.next;
curr = tail;
}
}
public void add(String item) {
if (curr != null) {
Node newNode = new Node(item, curr.next);
curr.next = newNode;
curr = newNode;
} else {
head = tail = new Node(item, null);
curr = head;
}
}
public void delete() {
if (curr.next == null) {
Node temp = head;
while (temp.next != curr) {
System.out.println(temp.item);
temp = temp.next;
}
temp.next = null;
curr = head;
}
}
public void find(String item) {
Node temp = new Node(curr.item, curr.next);
if (item.equals(temp.item))
curr = temp;
else {
temp = temp.next;
while (temp.next != null && temp != curr) {
if (item.equals(temp.item))
curr = temp;
}
}
}
public String get() {
if (curr != null)
return curr.item;
else
return "";
}
public boolean next() {
if (curr != tail) {
curr = curr.next;
return true;
} else
return false;
}
public void start() {
curr = head;
}
public void end() {
curr = tail;
}
public boolean empty() {
if (head == null)
return true;
else
return false;
}
}
Node class:
class Node {
Node next;
String item;
Node(String item, Node next) {
this.next = next;
this.item = item;
}
}
There is indeed a problem in add: it doesn't update tail when nodes already exist. Consider this sequence of actions:
LinkedList list = new LinkedList();
list.add("one");
list.add("two");
list.append("three");
If you were to then print it using this:
public void print() {
Node curr = this.head;
while(curr != null) {
System.out.println(curr.item);
curr = curr.next;
}
}
Like this:
list.print();
You'd get the following output:
one
three
This happens because tail -- which append relies on -- continues to point to the first Node after the second add operation is performed.
I don't see any problems here, so I would guess the issue is elsewhere.
Okay, the only issue I see there is in delete:
public void delete()
{
Node temp = head;
while(temp != null && temp.next != curr) {
System.out.println(temp.item);
temp=temp.next;
}
if (temp != null && temp.next != null) {
temp.next = temp.next.next;
}
curr = head;
}
I think I have found your problem.
If you use append() you add it directly after the tail. But when you have added previous nodes after the tail you don't set your tail to the new node. This means that once you call append() twice you loose all the nodes that you have added after the first append().
Brief example:
public static void main(String[] args) {
LinkedList list = new LinkedList();
list.add("First add");
list.append("First Append");
list.add("Second add");
list.prepend("First prepend");
list.add("Third add");
list.prepend("Second prepend");
list.add("fourth add");
list.append("Second Append");
list.add("Fifth add");
list.add("Sixth add");
list.start();
do {
System.out.println(list.get().toString());
} while (list.next());
}
Output:
Second prepend
fourth add
First prepend
Third add
First add
First Append
Second Append
Conclusion: "Second Add" is lost, as well as "Fifth add" and "Sixth add" because your next() method stops as soon as it reaches the tail. You need to always update the tail if you add a new node in the end.
Hope this helps.
Cheers, Chnoch
I think the problem is
if (curr != null) {
Node newNode = new Node(item, curr.next); //<-- here (curr.next)
//and
Node(String item, Node next) {
this.next = next; //<-- here
Try (Edited):
Node newNode = new Node(item, curr); // pass curr to the constructor of Node
curr = newNode;
I have been working on a Java project for a class for a while now. It is an implementation of a linked list (here called AddressList, containing simple nodes called ListNode). The catch is that everything would have to be done with recursive algorithms. I was able to do everything fine sans one method: public AddressList reverse()
ListNode:
public class ListNode{
public String data;
public ListNode next;
}
Right now my reverse function just calls a helper function that takes an argument to allow recursion.
public AddressList reverse(){
return new AddressList(this.reverse(this.head));
}
With my helper function having the signature of private ListNode reverse(ListNode current).
At the moment, I have it working iteratively using a stack, but this is not what the specification requires. I had found an algorithm in C that recursively reversed and converted it to Java code by hand, and it worked, but I had no understanding of it.
Edit: Nevermind, I figured it out in the meantime.
private AddressList reverse(ListNode current, AddressList reversedList){
if(current == null)
return reversedList;
reversedList.addToFront(current.getData());
return this.reverse(current.getNext(), reversedList);
}
While I'm here, does anyone see any problems with this route?
There's code in one reply that spells it out, but you might find it easier to start from the bottom up, by asking and answering tiny questions (this is the approach in The Little Lisper):
What is the reverse of null (the empty list)? null.
What is the reverse of a one element list? the element.
What is the reverse of an n element list? the reverse of the rest of the list followed by the first element.
public ListNode Reverse(ListNode list)
{
if (list == null) return null; // first question
if (list.next == null) return list; // second question
// third question - in Lisp this is easy, but we don't have cons
// so we grab the second element (which will be the last after we reverse it)
ListNode secondElem = list.next;
// bug fix - need to unlink list from the rest or you will get a cycle
list.next = null;
// then we reverse everything from the second element on
ListNode reverseRest = Reverse(secondElem);
// then we join the two lists
secondElem.next = list;
return reverseRest;
}
I was asked this question at an interview and was annoyed that I fumbled with it since I was a little nervous.
This should reverse a singly linked list, called with reverse(head,NULL);
so if this were your list:
1->2->3->4->5->null
it would become:
5->4->3->2->1->null
//Takes as parameters a node in a linked list, and p, the previous node in that list
//returns the head of the new list
Node reverse(Node n,Node p){
if(n==null) return null;
if(n.next==null){ //if this is the end of the list, then this is the new head
n.next=p;
return n;
}
Node r=reverse(n.next,n); //call reverse for the next node,
//using yourself as the previous node
n.next=p; //Set your next node to be the previous node
return r; //Return the head of the new list
}
edit: ive done like 6 edits on this, showing that it's still a little tricky for me lol
I got half way through (till null, and one node as suggested by plinth), but lost track after making recursive call. However, after reading the post by plinth, here is what I came up with:
Node reverse(Node head) {
// if head is null or only one node, it's reverse of itself.
if ( (head==null) || (head.next == null) ) return head;
// reverse the sub-list leaving the head node.
Node reverse = reverse(head.next);
// head.next still points to the last element of reversed sub-list.
// so move the head to end.
head.next.next = head;
// point last node to nil, (get rid of cycles)
head.next = null;
return reverse;
}
Here's yet another recursive solution. It has less code within the recursive function than some of the others, so it might be a little faster. This is C# but I believe Java would be very similar.
class Node<T>
{
Node<T> next;
public T data;
}
class LinkedList<T>
{
Node<T> head = null;
public void Reverse()
{
if (head != null)
head = RecursiveReverse(null, head);
}
private Node<T> RecursiveReverse(Node<T> prev, Node<T> curr)
{
Node<T> next = curr.next;
curr.next = prev;
return (next == null) ? curr : RecursiveReverse(curr, next);
}
}
The algo will need to work on the following model,
keep track of the head
Recurse till end of linklist
Reverse linkage
Structure:
Head
|
1-->2-->3-->4-->N-->null
null-->1-->2-->3-->4-->N<--null
null-->1-->2-->3-->4<--N<--null
null-->1-->2-->3<--4<--N<--null
null-->1-->2<--3<--4<--N<--null
null-->1<--2<--3<--4<--N<--null
null<--1<--2<--3<--4<--N
|
Head
Code:
public ListNode reverse(ListNode toBeNextNode, ListNode currentNode)
{
ListNode currentHead = currentNode; // keep track of the head
if ((currentNode==null ||currentNode.next==null )&& toBeNextNode ==null)return currentHead; // ignore for size 0 & 1
if (currentNode.next!=null)currentHead = reverse(currentNode, currentNode.next); // travarse till end recursively
currentNode.next = toBeNextNode; // reverse link
return currentHead;
}
Output:
head-->12345
head-->54321
I think this is more cleaner solution, which resembles LISP
// Example:
// reverse0(1->2->3, null) =>
// reverse0(2->3, 1) =>
// reverse0(3, 2->1) => reverse0(null, 3->2->1)
// once the first argument is null, return the second arg
// which is nothing but the reveresed list.
Link reverse0(Link f, Link n) {
if (f != null) {
Link t = new Link(f.data1, f.data2);
t.nextLink = n;
f = f.nextLink; // assuming first had n elements before,
// now it has (n-1) elements
reverse0(f, t);
}
return n;
}
I know this is an old post, but most of the answers are not tail recursive i.e. they do some operations after returning from the recursive call, and hence not the most efficient.
Here is a tail recursive version:
public Node reverse(Node previous, Node current) {
if(previous == null)
return null;
if(previous.equals(head))
previous.setNext(null);
if(current == null) { // end of list
head = previous;
return head;
} else {
Node temp = current.getNext();
current.setNext(previous);
reverse(current, temp);
}
return null; //should never reach here.
}
Call with:
Node newHead = reverse(head, head.getNext());
void reverse(node1,node2){
if(node1.next!=null)
reverse(node1.next,node1);
node1.next=node2;
}
call this method as reverse(start,null);
public Node reverseListRecursive(Node curr)
{
if(curr == null){//Base case
return head;
}
else{
(reverseListRecursive(curr.next)).next = (curr);
}
return curr;
}
public void reverse() {
head = reverseNodes(null, head);
}
private Node reverseNodes(Node prevNode, Node currentNode) {
if (currentNode == null)
return prevNode;
Node nextNode = currentNode.next;
currentNode.next = prevNode;
return reverseNodes(currentNode, nextNode);
}
Reverse by recursive algo.
public ListNode reverse(ListNode head) {
if (head == null || head.next == null) return head;
ListNode rHead = reverse(head.next);
rHead.next = head;
head = null;
return rHead;
}
By iterative
public ListNode reverse(ListNode head) {
if (head == null || head.next == null) return head;
ListNode prev = null;
ListNode cur = head
ListNode next = head.next;
while (next != null) {
cur.next = prev;
prev = cur;
cur = next;
next = next.next;
}
return cur;
}
public static ListNode recRev(ListNode curr){
if(curr.next == null){
return curr;
}
ListNode head = recRev(curr.next);
curr.next.next = curr;
curr.next = null;
// propogate the head value
return head;
}
This solution demonstrates that no arguments are required.
/**
* Reverse the list
* #return reference to the new list head
*/
public LinkNode reverse() {
if (next == null) {
return this; // Return the old tail of the list as the new head
}
LinkNode oldTail = next.reverse(); // Recurse to find the old tail
next.next = this; // The old next node now points back to this node
next = null; // Make sure old head has no next
return oldTail; // Return the old tail all the way back to the top
}
Here is the supporting code, to demonstrate that this works:
public class LinkNode {
private char name;
private LinkNode next;
/**
* Return a linked list of nodes, whose names are characters from the given string
* #param str node names
*/
public LinkNode(String str) {
if ((str == null) || (str.length() == 0)) {
throw new IllegalArgumentException("LinkNode constructor arg: " + str);
}
name = str.charAt(0);
if (str.length() > 1) {
next = new LinkNode(str.substring(1));
}
}
public String toString() {
return name + ((next == null) ? "" : next.toString());
}
public static void main(String[] args) {
LinkNode head = new LinkNode("abc");
System.out.println(head);
System.out.println(head.reverse());
}
}
Here is a simple iterative approach:
public static Node reverse(Node root) {
if (root == null || root.next == null) {
return root;
}
Node curr, prev, next;
curr = root; prev = next = null;
while (curr != null) {
next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
return prev;
}
And here is a recursive approach:
public static Node reverseR(Node node) {
if (node == null || node.next == null) {
return node;
}
Node next = node.next;
node.next = null;
Node remaining = reverseR(next);
next.next = node;
return remaining;
}
As Java is always pass-by-value, to recursively reverse a linked list in Java, make sure to return the "new head"(the head node after reversion) at the end of the recursion.
static ListNode reverseR(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode first = head;
ListNode rest = head.next;
// reverse the rest of the list recursively
head = reverseR(rest);
// fix the first node after recursion
first.next.next = first;
first.next = null;
return head;
}
PointZeroTwo has got elegant answer & the same in Java ...
public void reverseList(){
if(head!=null){
head = reverseListNodes(null , head);
}
}
private Node reverseListNodes(Node parent , Node child ){
Node next = child.next;
child.next = parent;
return (next==null)?child:reverseListNodes(child, next);
}
public class Singlelinkedlist {
public static void main(String[] args) {
Elem list = new Elem();
Reverse(list); //list is populate some where or some how
}
//this is the part you should be concerned with the function/Method has only 3 lines
public static void Reverse(Elem e){
if (e!=null)
if(e.next !=null )
Reverse(e.next);
//System.out.println(e.data);
}
}
class Elem {
public Elem next; // Link to next element in the list.
public String data; // Reference to the data.
}
public Node reverseRec(Node prev, Node curr) {
if (curr == null) return null;
if (curr.next == null) {
curr.next = prev;
return curr;
} else {
Node temp = curr.next;
curr.next = prev;
return reverseRec(curr, temp);
}
}
call using: head = reverseRec(null, head);
What other guys done , in other post is a game of content, what i did is a game of linkedlist, it reverse the LinkedList's member not reverse of a Value of members.
Public LinkedList reverse(LinkedList List)
{
if(List == null)
return null;
if(List.next() == null)
return List;
LinkedList temp = this.reverse( List.next() );
return temp.setNext( List );
}
package com.mypackage;
class list{
node first;
node last;
list(){
first=null;
last=null;
}
/*returns true if first is null*/
public boolean isEmpty(){
return first==null;
}
/*Method for insertion*/
public void insert(int value){
if(isEmpty()){
first=last=new node(value);
last.next=null;
}
else{
node temp=new node(value);
last.next=temp;
last=temp;
last.next=null;
}
}
/*simple traversal from beginning*/
public void traverse(){
node t=first;
while(!isEmpty() && t!=null){
t.printval();
t= t.next;
}
}
/*static method for creating a reversed linked list*/
public static void reverse(node n,list l1){
if(n.next!=null)
reverse(n.next,l1);/*will traverse to the very end*/
l1.insert(n.value);/*every stack frame will do insertion now*/
}
/*private inner class node*/
private class node{
int value;
node next;
node(int value){
this.value=value;
}
void printval(){
System.out.print(value+" ");
}
}
}
The solution is:
package basic;
import custom.ds.nodes.Node;
public class RevLinkedList {
private static Node<Integer> first = null;
public static void main(String[] args) {
Node<Integer> f = new Node<Integer>();
Node<Integer> s = new Node<Integer>();
Node<Integer> t = new Node<Integer>();
Node<Integer> fo = new Node<Integer>();
f.setNext(s);
s.setNext(t);
t.setNext(fo);
fo.setNext(null);
f.setItem(1);
s.setItem(2);
t.setItem(3);
fo.setItem(4);
Node<Integer> curr = f;
display(curr);
revLL(null, f);
display(first);
}
public static void display(Node<Integer> curr) {
while (curr.getNext() != null) {
System.out.println(curr.getItem());
System.out.println(curr.getNext());
curr = curr.getNext();
}
}
public static void revLL(Node<Integer> pn, Node<Integer> cn) {
while (cn.getNext() != null) {
revLL(cn, cn.getNext());
break;
}
if (cn.getNext() == null) {
first = cn;
}
cn.setNext(pn);
}
}
static void reverseList(){
if(head!=null||head.next!=null){
ListNode tail=head;//head points to tail
ListNode Second=head.next;
ListNode Third=Second.next;
tail.next=null;//tail previous head is poiniting null
Second.next=tail;
ListNode current=Third;
ListNode prev=Second;
if(Third.next!=null){
while(current!=null){
ListNode next=current.next;
current.next=prev;
prev=current;
current=next;
}
}
head=prev;//new head
}
}
class ListNode{
public int data;
public ListNode next;
public int getData() {
return data;
}
public ListNode(int data) {
super();
this.data = data;
this.next=null;
}
public ListNode(int data, ListNode next) {
super();
this.data = data;
this.next = next;
}
public void setData(int data) {
this.data = data;
}
public ListNode getNext() {
return next;
}
public void setNext(ListNode next) {
this.next = next;
}
}
private Node ReverseList(Node current, Node previous)
{
if (current == null) return null;
Node originalNext = current.next;
current.next = previous;
if (originalNext == null) return current;
return ReverseList(originalNext, current);
}
//this function reverses the linked list
public Node reverseList(Node p) {
if(head == null){
return null;
}
//make the last node as head
if(p.next == null){
head.next = null;
head = p;
return p;
}
//traverse to the last node, then reverse the pointers by assigning the 2nd last node to last node and so on..
return reverseList(p.next).next = p;
}
//Recursive solution
class SLL
{
int data;
SLL next;
}
SLL reverse(SLL head)
{
//base case - 0 or 1 elements
if(head == null || head.next == null) return head;
SLL temp = reverse(head.next);
head.next.next = head;
head.next = null;
return temp;
}
Inspired by an article discussing immutable implementations of recursive data structures I put an alternate solution together using Swift.
The leading answer documents solution by highlighting the following topics:
What is the reverse of nil (the empty list)?
Does not matter here, because we have nil protection in Swift.
What is the reverse of a one element list?
The element itself
What is the reverse of an n element list?
The reverse of the second element on followed by the first element.
I have called these out where applicable in the solution below.
/**
Node is a class that stores an arbitrary value of generic type T
and a pointer to another Node of the same time. This is a recursive
data structure representative of a member of a unidirectional linked
list.
*/
public class Node<T> {
public let value: T
public let next: Node<T>?
public init(value: T, next: Node<T>?) {
self.value = value
self.next = next
}
public func reversedList() -> Node<T> {
if let next = self.next {
// 3. The reverse of the second element on followed by the first element.
return next.reversedList() + value
} else {
// 2. Reverse of a one element list is itself
return self
}
}
}
/**
#return Returns a newly created Node consisting of the lhs list appended with rhs value.
*/
public func +<T>(lhs: Node<T>, rhs: T) -> Node<T> {
let tail: Node<T>?
if let next = lhs.next {
// The new tail is created recursively, as long as there is a next node.
tail = next + rhs
} else {
// If there is not a next node, create a new tail node to append
tail = Node<T>(value: rhs, next: nil)
}
// Return a newly created Node consisting of the lhs list appended with rhs value.
return Node<T>(value: lhs.value, next: tail)
}
Reversing the linked list using recursion. The idea is adjusting the links by reversing the links.
public ListNode reverseR(ListNode p) {
//Base condition, Once you reach the last node,return p
if (p == null || p.next == null) {
return p;
}
//Go on making the recursive call till reach the last node,now head points to the last node
ListNode head = reverseR(p.next); //Head points to the last node
//Here, p points to the last but one node(previous node), q points to the last node. Then next next step is to adjust the links
ListNode q = p.next;
//Last node link points to the P (last but one node)
q.next = p;
//Set the last but node (previous node) next to null
p.next = null;
return head; //Head points to the last node
}
public void reverseLinkedList(Node node){
if(node==null){
return;
}
reverseLinkedList(node.next);
Node temp = node.next;
node.next=node.prev;
node.prev=temp;
return;
}
public void reverse(){
if(isEmpty()){
return;
}
Node<T> revHead = new Node<T>();
this.reverse(head.next, revHead);
this.head = revHead;
}
private Node<T> reverse(Node<T> node, Node<T> revHead){
if(node.next == null){
revHead.next = node;
return node;
}
Node<T> reverse = this.reverse(node.next, revHead);
reverse.next = node;
node.next = null;
return node;
}
Here is a reference if someone is looking for Scala implementation:
scala> import scala.collection.mutable.LinkedList
import scala.collection.mutable.LinkedList
scala> def reverseLinkedList[A](ll: LinkedList[A]): LinkedList[A] =
ll.foldLeft(LinkedList.empty[A])((accumulator, nextElement) => nextElement +: accumulator)
reverseLinkedList: [A](ll: scala.collection.mutable.LinkedList[A])scala.collection.mutable.LinkedList[A]
scala> reverseLinkedList(LinkedList("a", "b", "c"))
res0: scala.collection.mutable.LinkedList[java.lang.String] = LinkedList(c, b, a)
scala> reverseLinkedList(LinkedList("1", "2", "3"))
res1: scala.collection.mutable.LinkedList[java.lang.String] = LinkedList(3, 2, 1)