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JAVA: My Linkedlist is only printing my head element and nothing else

I have written my own linked list and am reading integers from a file into the list and printing them out. However, only the head of my list is printing and nothing else. I've been staring at this code for so long I feel insane, can anyone help?
Method in a separate 'files' class that reads in a file of integers separated by whitespace. This method will take the next integer and add it to my linked list.
public void readValues() {
LinkedList list = new LinkedList();
while(scan.hasNextInt()) {
Integer someData = scan.nextInt();
list.addNode(someData);
}
list.printList();
}
This method is in my LinkedList class which takes the data sent from my readValues method in my files class.
public void addNode(Integer someData) {
myNode = new LinkedNode(someData,null);
//initialize node if this is first element
if (head == null) {
head = myNode;
size++;
}
else if (myNode.getNext() == null) {
myNode.setNext(myNode);
size ++;
}
else if (myNode.getNext() != null) {
while(myNode.getNext() != null) {
myNode = myNode.getNext();
}
myNode.setNext(myNode);
size++;
}
}
This method is also in my LinkedList class and successfully prints the head of my list which with my data is the number 40 followed by ---> and then nothing else. It should print ever other integer read in from my file.
public void printList() {
LinkedNode current = head;
if (head == null) {
System.out.print("list is empty");
return;
}
while(current != null) {
System.out.print(current.getElement());
System.out.print(" --> ");
current = current.getNext();
}
}
LinkedNode class:
public class LinkedNode {
Integer data;
LinkedNode next;
public LinkedNode(Integer someData, LinkedNode next) {
this.data = someData;
this.next = null;
}
public int getElement() {
return data;
}
public void setElement(Integer data) {
this.data = data;
}
public LinkedNode getNext() {
return next;
}
public void setNext(LinkedNode next) {
this.next = next;
}
public String toString() {
return data + "";
}
}

Your code has a small bug in the if else conditions in your addNode() method due to which your data is not getting added in the list.
Root Cause
When you add a new node to your list,
In the first iteration
The head is currently null and hence the first if conditions becomes true and your first node gets added (That's why you got the data 40).
In the Subsequent iteration(s)
your else if condition checks the myNode's next pointer which will always be null (as per the constructor) and thus it's next pointer points towards itself. The nodes created from here do not become the part of the list as the next pointer of head was never assigned to any of these and these nodes also point to themselves only.
Solution
I made a little modification in the if else conditions:
public void addNode(Integer someData) {
LinkedNode myNode = new LinkedNode(someData,null);
//initialize node if this is first element
if (head == null) {
head = myNode;
size++;
}
else if (head.getNext() == null) {
head.setNext(myNode);
size ++;
}
else if (head.getNext() != null) {
System.out.println("in second else if");
LinkedNode n = head;
while(n.getNext() != null) {
n = n.getNext();
}
n.setNext(myNode);
size++;
}
}
PS: Try debugging your code with dry run, it's a great mental exercise and helps in boosting the learning curve significantly too. All the best! :)

The problem is with your addNode() method. Inside the addNode() method you are first creating a new node named mynode. Now when the head is null it sets head to mynode, thats ok. But when the head is not null the mynode is not being added to the list. Thats why only the first element exist and other's are getting lost.
Hope this helps. Let me know if I can help with anything else. Happy coding!

Segregate even and odd nodes in a Linked List

Given, a linked-list, I'm trying to partition it into so that the even nodes come before the odd nodes. My approach is to create two different linked-list (even and odd) to store even numbers and odd numbers. However, I'm running into a problem when I want to add to the even or odd linked list (I commented the part that I think is giving me problem in my code below). Thanks!
public class SeperateOddEven {
static Node head;
static int count;
public static class Node {
int data;
Node next;
private Node(int data) {
this.data = data;
next = null;
count++;
}
}
public void seperate() {
Node even = null;
Node odd = null;
Node temp;
// go through each linked-list and place node in new list depending on whether they are even or odd
while(head != null) {
// if even, place in even linked-list
if(head.data % 2 == 0) {
temp = new Node(head.data);
even = temp; // Problem here
even = even.next; // and here
} else { // if head.data % 2 != 0
temp = new Node(head.data);
odd = temp;
odd = odd.next;
}
head = head.next;
}
toString(even);
//toString(odd);
}
public void toString(Node node) {
while (node != null) {
System.out.print(node.data + " ");
node = node.next;
}
}
public static void main(String[] args) {
SeperateOddEven s = new SeperateOddEven();
head = new Node(8);
head.next = new Node(12);
head.next.next = new Node(10);
head.next.next.next = new Node(5);
head.next.next.next.next = new Node(4);
head.next.next.next.next.next = new Node(1);
head.next.next.next.next.next.next = new Node(6);
System.out.println("original list: ");
s.toString(head);
s.seperate();
}
}

I believe you identified exactly where the problem is. Let's go line by line:
temp = new Node(head.data);
The extra temp variable is unnecessary but fine.
even = temp;
A problem arises on the next line however. You assign even to temp (making temp unnecessary). If something was previously stored in even, it is now lost to the garbage collector because you now have no reference to it. even and temp are now both references to the same Node object.
What I think you might have wanted to do was to say even.next = temp. This would start to create a list, but with only a single reference you would have to use that reference to point to the head of the list. Each time you wanted to append to the list, you would need to loop through it until you found the end. If you instead tried to make this single reference point to the tail of the list, you would no longer have any way to get back to the head because your Nodes only have next references, and not prev references (a list with bidirectional references is called a doubly linked list).
even = even.next;
Because even (and temp) both point to the newly created Node object, the even.next property is null. So when this line executes, even now points to null. The work inside the loop has accomplished nothing because you immediately lose references to every Node you create.
Try something like this:
// Must keep track of head reference, because your Nodes can only go forward
Node evenHead = null;
Node evenTail = null;
Node oddHead = null;
Node oddTail = null;
while (head != null) {
if(head.data % 2 == 0) {
if (evenHead == null) {
// The even list is empty, set the head and tail
evenHead = new Node(head.data);
evenTail = evenHead;
} else {
// Append to the end of the even list
evenTail.next = new Node(head.data);
evenTail = evenTail.next;
}
} else {
// similar code for odd, consider creating a method to avoid repetition
}
}

You can also try this :
while (head != null) {
// if even, place in even linked-list
temp = new Node(head.data);
if (head.data % 2 == 0) {
if(even == null) {
even = temp;
} else{
Node insertionNode = even;
while(insertionNode.next != null)
insertionNode = insertionNode.next;
insertionNode.next = temp;
}
} else { // if head.data % 2 != 0
if(odd == null) {
odd = temp;
} else{
Node insertionNode = odd;
while(insertionNode.next != null)
insertionNode = insertionNode.next;
insertionNode.next = temp;
}
}
head = head.next;
}

Java remove duplicates from linked list

I want to remove duplicates from sorted linked list {0 1 2 2 3 3 4 5}.
`
public Node removeDuplicates(Node header)
{
Node tempHeader = null;
if(header != null)
tempHeader = header.next;
else return header;
Node prev = header;
if((tempHeader == null)) return header ;
while(tempHeader != null)
{
if(tempHeader.data != prev.data)
{
prev.setNext(tempHeader);
}
tempHeader = tempHeader.next;
}
prev = header;
printList(prev);
return tempHeader;
}
`
prev.setNext(tempHeader) is not working correctly inside the while loop. Ideally when prev = 2 and tempHeader = 3, prev.next should be node with data = 3.
Printlist function just takes header pointer and prints the list.
Node definition is given below.
public class Node
{
int data;
Node next;
public Node getNext() {
return next;
}
public void setNext(Node next) {
this.next = next;
}
}

The loop is sorted, so you know that duplicates are going to sit next to each other. If you want to edit the list in place then, you've got to have two list pointers (which you do). The one you call tempHeader and prev, and you've got to advance them both in the the list as you go (which I don't see in the code). Otherwise, if you don't advance the prev pointer as you go, then you're always comparing the element under tempHeader to the first item in the list, which is not correct.
An easier way to do this, however, is to build a new list as you go. Simply remember the value of the last item that you appended to the list. Then if the one that you're about to insert is the same then simply don't insert it, and when you're done, just return your new list.

I can give you 2 suggestions for the above suggestion
1) Convert the linked List to Set, that will eliminate the duplicates and
Back from Set to the Linked list
Code to get this done would be
linkedList = new LinkedList<anything>(new HashSet<anything>(origList));
2) You can use LinkedHashSet, if you dont want any duplicates

In this case no return value is needed.
public void removeDuplicates(Node list) {
while (list != null) {
// Walk to next unequal node:
Node current = list.next;
while (current != null && current.data.equals(list.data)) {
current = current.next;
}
// Skip the equal nodes:
list.next = current;
// Take the next unequal node:
list = current;
}
}

public ListNode removeDuplicateElements(ListNode head) {
if (head == null || head.next == null) {
return null;
}
if (head.data.equals(head.next.data)) {
ListNode next_next = head.next.next;
head.next = null;
head.next = next_next;
removeDuplicateElements(head);
} else {
removeDuplicateElements(head.next);
}
return head;
}

By DoublyLinked List and using HashSet,
public static void deleteDups(Node n) {
HashSet<Integer> set = new HashSet<Integer>();
Node previous = null;
while (n != null) {
if (set.contains(n.data)) {
previous.next = n.next;
} else {
set.add(n.data);
previous = n;
}
n = n.next;
}
}
doublylinkedList
class Node{
public Node next;
public Node prev;
public Node last;
public int data;
public Node (int d, Node n, Node p) {
data = d;
setNext(n);
setPrevious(p);
}
public Node() { }
public void setNext(Node n) {
next = n;
if (this == last) {
last = n;
}
if (n != null && n.prev != this) {
n.setPrevious(this);
}
}
public void setPrevious(Node p) {
prev = p;
if (p != null && p.next != this) {
p.setNext(this);
}
}}

Invert linear linked list

a linear linked list is a set of nodes. This is how a node is defined (to keep it easy we do not distinguish between node an list):
class Node{
Object data;
Node link;
public Node(Object pData, Node pLink){
this.data = pData;
this.link = pLink;
}
public String toString(){
if(this.link != null){
return this.data.toString() + this.link.toString();
}else{
return this.data.toString() ;
}
}
public void inc(){
this.data = new Integer((Integer)this.data + 1);
}
public void lappend(Node list){
Node child = this.link;
while(child != null){
child = child.link;
}
child.link = list;
}
public Node copy(){
if(this.link != null){
return new Node(new Integer((Integer)this.data), this.link.copy());
}else{
return new Node(new Integer((Integer)this.data), null);
}
}
public Node invert(){
Node child = this.link;
while(child != null){
child = child.link;
}
child.link = this;....
}
}
I am able to make a deep copy of the list. Now I want to invert the list so that the first node is the last and the last the first. The inverted list has to be a deep copy.
I started developing the invert function but I am not sure. Any Ideas?
Update: Maybe there is a recursive way since the linear linked list is a recursive data structure.
I would take the first element, iterate through the list until I get to a node that has no child and append the first element, I would repeat this for the second, third....

I sometimes ask this question in interviews...
I would not recommend using a recursive solution, or using a stack to solve this. There's no point in allocating O(n) memory for such a task.
Here's a simple O(1) solution (I didn't run it right now, so I apologize if it needs some correction).
Node reverse (Node current) {
Node prev = null;
while (current != null) {
Node nextNode = current.next;
current.next = prev;
prev = current;
current = nextNode;
}
return prev;
}
BTW: Does the lappend method works? It seems like it would always throw a NullReferenceException.

There's a great recursive solution to this problem based on the following observations:
The reverse of the empty list is the empty list.
The reverse of a singleton list is itself.
The reverse of a list of a node N followed by a list L is the reverse of the list L followed by the node N.
You can therefore implement the reverse function using pseudocode along these lines:
void reverseList(Node node) {
if (node == null) return; // Reverse of empty list is itself.
if (node.next == null) return; // Reverse of singleton list is itself.
reverseList(node.next); // Reverse the rest of the list
appendNodeToList(node, node.next); // Append the new value.
}
A naive implementation of this algorithm runs in O(n2), since each reversal requires an append, which requires an O(n) scan over the rest of the list. However, you can actually get this working in O(n) using a clever observation. Suppose that you have a linked list that looks like this:
n1 --> n2 --> [rest of the list]
If you reverse the list beginning at n2, then you end up with this setup:
n1 [reverse of rest of the list] --> n2
| ^
+------------------------------------------+
So you can append n1 to the reverse of the rest of the list by setting n1.next.next = n1, which changes n2, the new end of the reverse list, to point at n1:
[reverse of the rest of the list] --> n2 --> n1
And you're golden! Again more pseudocode:
void reverseList(Node node) {
if (node == null) return; // Reverse of empty list is itself.
if (node.next == null) return; // Reverse of singleton list is itself.
reverseList(node.next); // Reverse the rest of the list
node.next.next = node; // Append the new value.
}
EDIT: As Ran pointed out, this uses the call stack for its storage space and thus risks a stack overflow. If you want to use an explicit stack instead, you can do so like this:
void reverseList(Node node) {
/* Make a stack of the reverse of the nodes. */
Stack<Node> s = new Stack<Node>();
for (Node curr = node; node != null; node = node.next)
s.push(curr);
/* Start unwinding it. */
Node curr = null;
while (!s.empty()) {
Node top = s.pop();
/* If there is no node in the list yet, set it to the current node. */
if (curr == null)
curr = top;
/* Otherwise, have the current node point to this next node. */
else
curr.next = top;
/* Update the current pointer to be this new node. */
curr = top;
}
}
I believe that this similarly inverts the linked list elements.

I would treat the current list as a stack (here's my pseudo code):
Node x = copyOf(list.head);
x.link = null;
foreach(node in list){
Node temp = copyOf(list.head);
temp.link = x;
x = temp;
}
At the end x will be the head of the reversed list.

I more fammiliar whit C, but still let me try. ( I just do not sure if this runs in Java, but it should)
node n = (well first one)
node prev = NULL;
node t;
while(n != NULL)
{
t = n.next;
n.next = prev;
prev = n;
n = t;
}

Reversing a single-linked list is sort of a classic question. It's answered here as well (and well answered), it does not requires recursion nor extra memory, besides a register (or 2) for reference keeping.
However to the OP, I guess it's a school project/homework and some piece of advice, if you ever get to use single linked list for some real data storage, consider using a tail node as well. (as of now single linked lists are almost extinct, HashMap buckets comes to mind, though).
Unless you have to check all the nodes for some condition during 'add', tail is quite an improvement. Below there is some code that features the reverse method and a tail node.
package t1;
public class SList {
Node head = new Node();
Node tail = head;
private static class Node{
Node link;
int data;
}
void add(int i){
Node n = new Node();
n.data = i;
tail = tail.link =n;
}
void reverse(){
tail = head;
head = reverse(head);
tail.link = null;//former head still links back, so clear it
}
private static Node reverse(Node head){
for (Node n=head.link, link; n!=null; n=link){//essentially replace head w/ the next and relink
link = n.link;
n.link = head;
head = n;
}
return head;
}
void print(){
for (Node n=head; n!=null;n=n.link){
System.out.println(n.data);
}
}
public static void main(String[] args) {
SList l = new SList();
l.add(1);l.add(2);l.add(3);l.add(4);
l.print();
System.out.println("==");
l.reverse();
l.print();
}
}

I was wondering something like that(I didnt test it, so):
invert(){
m(firstNode, null);
}
m(Node v, Node bef){
if(v.link != null)
m(v.link,v);
else
v.link=bef;
}

Without much testing,
Node head = this;
Node middle = null;
Node trail = null;
while (head != null) {
trail = middle;
middle = head;
head = head.link;
middle.link = trail;
}
head = middle;
return head;

public ListNode Reverse(ListNode list)
{
if (list == null) return null;
if (list.next == null) return list;
ListNode secondElem = list.next;
ListNode reverseRest = Reverse(secondElem);
secondElem.Next = list;
return reverseRest;
}
Hope this helps.

Manually sorting a linked list in Java (lexically)

I am implementing my own linked list in Java. The node class merely has a string field called "name" and a node called "link". Right now I have a test driver class that only inserts several names sequentially. Now, I am trying to write a sorting method to order the nodes alphabetically, but am having a bit of trouble with it. I found this pseudocode of a bubblesort from someone else's post and tried to implement it, but it doesn't fully sort the entries. I'm not really quite sure why. Any suggestions are appreciated!
private void sort()
{
//Enter loop only if there are elements in list
boolean swapped = (head != null);
// Only continue loop if a swap is made
while (swapped)
{
swapped = false;
// Maintain pointers
Node curr = head;
Node next = curr.link;
Node prev = null;
// Cannot swap last element with its next
while (next != null)
{
// swap if items in wrong order
if (curr.name.compareTo(next.name) < 0)
{
// notify loop to do one more pass
swapped = true;
// swap elements (swapping head in special case
if (curr == head)
{
head = next;
Node temp = next.link;
next.link = curr;
curr.link = temp;
curr = head;
}
else
{
prev.link = curr.link;
curr.link = next.link;
next.link = curr;
curr = next;
}
}
// move to next element
prev = curr;
curr = curr.link;
next = curr.link;
}
}
}

I spent some minutes eyeballing your code for errors but found none.
I'd say until someone smarter or more hard working comes along you should try debugging this on your own. If you have an IDE like Eclipse you can single-step through the code while watching the variables' values; if not, you can insert print statements in a few places and hand-check what you see with what you expected.
UPDATE I
I copied your code and tested it. Apart from the fact that it sorts in descending order (which may not be what you intended) it worked perfectly for a sample of 0, 1 and 10 random nodes. So where's the problem?
UPDATE II
Still guessing what could be meant by "it doesn't fully sort the entries." It's possible that you're expecting lexicographic sorting (i.e. 'a' before 'B'), and that's not coming out as planned for words with mixed upper/lower case. The solution in this case is to use the String method compareToIgnoreCase(String str).

This may not be the solution you're looking for, but it's nice and simple. Maybe you're lazy like I am.
Since your nodes contain only a single item of data, you don't really need to re-shuffle your nodes; you could simply exchange the values on the nodes while leaving the list's structure itself undisturbed.
That way, you're free to implement Bubble Sort quite simply.

you should use the sorting procedures supplied by the language.
try this tutorial.
Basically, you need your element class to implement java.lang.Comparable, in which you will just delegate to obj.name.compareTo(other.name)
you can then use Collections.sort(yourCollection)
alternatively you can create a java.util.Comparator that knows how to compare your objects

To obtain good performance you can use Merge Sort.
Its time complexity is O(n*log(n)) and can be implemented without memory overhead for lists.
Bubble sort is not good sorting approach. You can read the What is a bubble sort good for? for details.

This may be a little too late. I would build the list by inserting everything in order to begin with because sorting a linked list is not fun.
I'm positive your teacher or professor doesn't want you using java's native library. However that being said, there is no real fast way to resort this list.
You could read all the nodes in the order that they are in and store them into an array. Sort the array and then relink the nodes back up. I think the Big-Oh complexity of this would be O(n^2) so in reality a bubble sort with a linked list is sufficient

I have done merge sort on the singly linked list and below is the code.
public class SortLinkedList {
public static Node sortLinkedList(Node node) {
if (node == null || node.next == null) {
return node;
}
Node fast = node;
Node mid = node;
Node midPrev = node;
while (fast != null && fast.next != null) {
fast = fast.next.next;
midPrev = mid;
mid = mid.next;
}
midPrev.next = null;
Node node1 = sortLinkedList(node);
Node node2 = sortLinkedList(mid);
Node result = mergeTwoSortedLinkedLists(node1, node2);
return result;
}
public static Node mergeTwoSortedLinkedLists(Node node1, Node node2) {
if (null == node1 && node2 != null) {
return node2;
} else if (null == node2 && node1 != null) {
return node1;
} else if (null == node1 && null == node2) {
return null;
} else {
Node result = node1.data <= node2.data ? node1 : node2;
Node prev1 = null;
while (node1 != null && node2 != null) {
if (node1.data <= node2.data) {
prev1 = node1;
node1 = node1.next;
} else {
Node next2 = node2.next;
node2.next = node1;
if (prev1 != null) {
prev1.next = node2;
}
node1 = node2;
node2 = next2;
}
}
if (node1 == null && node2 != null) {
prev1.next = node2;
}
return result;
}
}
public static void traverseNode(Node node) {
while (node != null) {
System.out.print(node + " ");
node = node.next;
}
System.out.println();
}
public static void main(String[] args) {
MyLinkedList ll1 = new MyLinkedList();
ll1.insertAtEnd(10);
ll1.insertAtEnd(2);
ll1.insertAtEnd(20);
ll1.insertAtEnd(4);
ll1.insertAtEnd(9);
ll1.insertAtEnd(7);
ll1.insertAtEnd(15);
ll1.insertAtEnd(-3);
System.out.print("list: ");
ll1.traverse();
System.out.println();
traverseNode(sortLinkedList(ll1.start));
}
}
The Node class:
public class Node {
int data;
Node next;
public Node() {
data = 0;
next = null;
}
public Node(int data) {
this.data = data;
}
public int getData() {
return this.data;
}
public Node getNext() {
return this.next;
}
public void setData(int data) {
this.data = data;
}
public void setNext(Node next) {
this.next = next;
}
#Override
public String toString() {
return "[ " + data + " ]";
}
}
The MyLinkedList class:
public class MyLinkedList {
Node start;
public void insertAtEnd(int data) {
Node newNode = new Node(data);
if (start == null) {
start = newNode;
return;
}
Node traverse = start;
while (traverse.getNext() != null) {
traverse = traverse.getNext();
}
traverse.setNext(newNode);
}
public void traverse() {
if (start == null)
System.out.println("List is empty");
else {
Node tempNode = start;
do {
System.out.print(tempNode.getData() + " ");
tempNode = tempNode.getNext();
} while (tempNode != null);
System.out.println();
}
}
}