Reputation: 39646
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?
Upvotes: 108
Views: 192719
Reputation: 113
Solution in javascript (recursive) :
function reverse_linked_list_1(node){
function reverse_linked_list_1(node, result){
return node ? reverse_linked_list_1(node.next, {data: node.data, next: result}) : result;
}
return reverse_linked_list_1(node, null);
}
Upvotes: -1
Reputation: 49219
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):
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;
}
Upvotes: 332
Reputation: 34424
public void reverseLinkedList(Node node){
if(node==null){
return;
}
reverseLinkedList(node.next);
Node temp = node.next;
node.next=node.prev;
node.prev=temp;
return;
}
Upvotes: 0
Reputation: 1166
Here is C# version of Reverse for linklist.
public void Reverse()
{
Node currentNode, nextNode=null, prevNode=null;
currentNode = head;
while(currentNode!=null)
{
nextNode = currentNode.next;
currentNode.next = prevNode;
prevNode = currentNode;
currentNode = nextNode;
}
head = prevNode;
}
Upvotes: -3
Reputation: 1329
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
}
Upvotes: 0
Reputation: 5064
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:
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)
}
Upvotes: 0
Reputation: 31
//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;
}
Upvotes: 0
Reputation: 7
This is how we would do this in Opal - a pure functional programming language. And, IMHO - doing this recursively only makes sense in that context.
List Reverse(List l)
{
if (IsEmpty(l) || Size(l) == 1) return l;
return reverse(rest(l))::first(l);
}
rest(l) returns a list that is the original list without it's first node. first(l) returns the first element. :: is a concatenation operator.
Upvotes: -2
Reputation: 11
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);
}
Upvotes: 1
Reputation: 1
//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;
}
Upvotes: 0
Reputation: 1779
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;
}
Upvotes: 2
Reputation: 1
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);
}
Upvotes: 0
Reputation: 191
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;
}
}
Upvotes: 0
Reputation: 1
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);
}
}
Upvotes: 0
Reputation: 2232
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
Upvotes: 29
Reputation: 710
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;
}
Upvotes: 2
Reputation: 21
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());
}
}
Upvotes: 2
Reputation: 2144
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)
Upvotes: -1
Reputation: 51
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);
}
Upvotes: 5
Reputation: 1
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;
}
Upvotes: -1
Reputation: 43
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;
}
Upvotes: 3
Reputation: 1
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+" ");
}
}
}
Upvotes: 0
Reputation: 423
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 );
}
Upvotes: 0
Reputation: 759
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);
Upvotes: 0
Reputation: 195
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());
Upvotes: 7
Reputation: 21
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;
}
Upvotes: 2
Reputation: 179
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.
}
Upvotes: 0
Reputation: 81
The algo will need to work on the following model,
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
Upvotes: 8
Reputation: 2413
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;
}
Upvotes: 7
Reputation: 2314
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);
}
}
Upvotes: 9