So I have a linked list that I am trying to remove duplicates from.
My basic algorithm that I thought up is to pretty much use the runner technique. Where I keep two pointers to compare adjacent elements. If they are the same I change the pointer of p1 to point to p1.next.next if not I keep traversing the list. However I keep getting a null pointer exception in the solution I have typed.
Node RemoveDuplicates(Node head) {
// This is a "method-only" submission.
// You only need to complete this method.
if (head == null){
return null;
}
Node current = head;
Node runner = head;
while(current != null && runner != null && runner.next != null){
runner = runner.next;
if(runner.data == current.data){
if(current.next != null){
current = current.next.next;
}
}else{
current = current.next;
}
}
return current;
}
At the point that I exit the while loop current is null. Which I think is the problem. How would I return the head of the altered list.
OK, although you've already accepted an answer, here's some example code using recursion to remove the dups from an ordered list per your request in the comments. (if your list isn't ordered, order it :) )
public Node removeDups(Node root) {
if (root.next == null)
return root;
root.next = removeDups(root.next);
if (root.data == root.next.data)
return root.next;
return root;
} // call as root = removeDups(root);
As you mentioned, recursion isn't really necessary here but you're using a Node-based linked list which is recursively defined. So, when it makes sense, the elegance of the solution has its benefits.
What I like about it is that you're not doing any node.next.next or needing to check for that null case. Once the stack starts unwinding, you're already in a position to start checking for dups. Then it's just a matter of comparing root.data and root.next.data; both of which you already know exist.
You can do this with single traversal with 2 pointer.and also this code works with single while loop.
public Node deleteDuplicates(Node head) {
Node current=head;
if (head == null)
return null;
else
{
Node runner=head.next;
while(head.next!=null && runner!=null)
{
if(head.val == runner.val)
prev=runner.next;
else
{
head.next=runner;
head=head.next;
prev=runner.next;
}
}
head.next=runner;
}
return current;
}
First of all, you'll want to return head at the end, so that you are returning the list, not just the last element.
Second thing, you'll want to modify the .next references instead of assigning them in some of the cases.
Note this doesn't work if the list isn't sorted.
Before: 1 1 3 3
After: 1 3
This code works (and I've tested it)
static Node RemoveDuplicates(Node head) {
if (head == null) return null;
Node current = head;
Node runner = head;
while (current != null && current.next != null) {
runner = current.next;
while (runner != null && runner.data == current.data) {
current.next = runner.next; // skip the repeat
runner = runner.next;
}
current = current.next;
}
return head;
}
You can do this in a single traversal. Just maintain two pointers temp
and next_of_next. Make temp iterate for each node and when the data of temp and the next node is equal, point next_of_next to the alternate node after temp and delete the node after temp.
Node removeDuplicates(Node head)
{
Node temp = head;
Node next_of_next;
if (head == null)
return;
while (temp.next != null)
{
if (temp.data == temp.next.data)
{
next_of_next = temp.next.next;
temp.next = null;
temp.next = next_of_next;
}
else
temp = temp.next;
}
return head ;
}
Here an apporach without recursion using a HashSet:
public void RemoveDuplicates()
{
if (head != null)
{
var hm = new HashSet<T>();
Node current = head;
Node prev = null;
while (current != null)
{
if (!hm.Contains(current.Value))
{
hm.Add(current.Value);
prev = current;
current = current.Next;
}
else
{
prev.Next = current.Next;
current = prev.Next;
}
}
}
return head;
}
Related
I'm having a problem deleting nodes from singly linked lists in Java. I have a list, which has data of integers, and I need to delete all nodes, whose value can be divided by divided by four. I also need to move the head and tail pointers in case either of the head or tail elements are deleted. I wrote method for this, and most of the time it works just like I need it to, but sometimes it throws NullPointerException. How can I fix it? Here's my code:
public void delete(){
Node temp = head, prev = null;
if (temp != null && temp.data % 4 == 0)
{
head = temp.next;
temp = head;
}
while (temp.next != null)
{
if (temp.data % 4 != 0) {
prev = temp;
temp = temp.next;
}
else {
prev.next = temp.next;
temp = prev.next;
}
}
if (tail.data % 4 == 0) {
tail = prev;
tail.next = null;
}
}
while (temp.next != null): temp may be null. And some more small problems.
This is due to too much complexity.
public void delete() {
Node prev = null;
for (Node temp = head; temp != null; temp = temp.next) {
if (temp.data % 4 == 0) {
if (prev == null) {
head = temp.next;
} else {
prev.next = temp.next;
}
} else {
prev = temp;
}
}
tail = prev;
}
The above sets prev to the valid previous node.
The deletion considers deletion from head or from prev.next.
The tail is updated to the last element.
in your while condition add one more null check:
while (null != temp && null != temp.next)
As I've just started programming a few months back a lot of new information is coming and I'm having trouble catching up.So here I have created what I thought was a sorted linked list.Turns out it is not sorted
public boolean insert(Person person) {
Node n = new Node(person);
Node p = head;
if(p == null) {
head = n;
size++;
return true;
} else {
Node temp = p;
int comparison;
while(temp.next != null) {
comparison = temp.person.name.compareTo(person.name);
if(comparison == 0){
return false;
}
temp = temp.next;
}
temp.next = n;
size++;
return true;
}
}
The method works,it inserts the persons,but they arent sorted like they should be.What part of the code do I need to change/remove in order to make it sort.
Thanks!
You should insert like this:
static boolean insert(Person person) {
Node newNode = new Node(person);
if (head == null) {
head = newNode;
size++;
return true;
}
Node current = head;
Node prev = null;
int comparison;
while (current != null) {
comparison = person.name.compareTo(current.person.name);
if (comparison == 0) {
return false;
} else if (comparison > 0) { /// greater than
if (current.next == null) { // check if reach tail of the linked list add and break
current.next = newNode;
break;
}
} else { // less then
if (prev == null) { // check if it should be first then put and break
Node oldHead = head;
head = newNode;
head.next = oldHead;
break;
}
prev.next = newNode;
newNode.next = current;
break;
}
prev = current;
current = current.next;
}
size++;
return true;
}
There is a problem in your else part. You are returning false when same value is given. But it is not interpreted properly for a valid case.
You need to have as below.
Check current node value - Check for null pointer exception
Check next node value - Check for null pointer exception
If current input is between currentNode and nextNode then do insert between.
If reaches last node, then insert at the end
I'm on HackerRank and I need to remove duplicate items from a sorted linked list. I passed all the cases except for two of them which the input is something like: 10001102034
So my program takes to seconds to complete and exceed the time. How can I do my code more efficiently, I heard about using square root but I don't know how to use it. Any guide is appreciate. Here is my code.
private static Node removeDuplicates(Node head) {
/* Another reference to head */
Node current = head;
Node next;
/* Traverse list till the last node */
while (current != null && current.next != null) {
if (current.data == current.next.data) {
next = current.next.next;
if (next == null) {
current.next = null;
break;
}
current.next = next;
} else {
current = current.next;
}
}
return head;
}
Again. It works but takes too much times with longer numbers.
You should replace condition if (current.data == current.next.data) with while loop and use break 'label':
out:
while (current != null && current.next != null) {
while (current.data == current.next.data) {
next = current.next.next;
if (next == null) {
current.next = null;
break out;
}
current.next = next;
}
current = current.next;
}
You can't use the square root because when u want to remove duplicates from a list you have to check all the list .
The square root technique is used for searching in a sorted list.
But for your question if you can improve the runtime on that your code in O(n^2) but if you change your code to use hashtable you can make it O(n).
import java.util.HashSet;
public class removeDuplicates
{
static class node
{
int val;
node next;
public node(int val)
{
this.val = val;
}
}
/* Function to remove duplicates from a
unsorted linked list */
static void removeDuplicate(node head)
{
// Hash to store seen values
HashSet<Integer> hs = new HashSet<>();
/* Pick elements one by one */
node current = head;
node prev = null;
while (current != null)
{
int curval = current.val;
// If current value is seen before
if (hs.contains(curval)) {
prev.next = current.next;
} else {
hs.add(curval);
prev = current;
}
current = current.next;
}
}
/* Function to print nodes in a given linked list */
static void printList(node head)
{
while (head != null)
{
System.out.print(head.val + " ");
head = head.next;
}
}
I hope this will help you.
I'm new to Java.
I have created a method where it will remove elements from LinkedList except the first one. The idea is a boolean will be set to true if a LinkedList's element data (Which is in Integer) matched with parameter. Once the boolean sets to true, it will remove any element that also matched with initial one.
Now for the problem. For example, if I were to remove 5 except the first one from this LinkedList:
5 5 5 6 5 7 8 9
I will get result like this:
5 5 6 7 8 9
As you can see, it didn't remove the 5 on the second position. Is there anything wrong with my code?
Here's the code by the way
public void append(int data) {
Node newNode = new Node(data);
if (head == null) {
head = new Node(data);
return;
}
Node lastNode = head;
while (lastNode.next != null) {
lastNode = lastNode.next;
}
lastNode.next = newNode;
return;
}
public void insert(int data) {
Node newData = new Node(data);
newData.next = head;
head = newData;
}
public void removeExceptFirst(int dataValue) { //The mentioned method
boolean duplicate = false;
Node currentNode = head;
while (currentNode.next != null) {
int value = currentNode.next.data;
if (value == dataValue) {
if (!duplicate) {
duplicate = true;
currentNode = currentNode.next;
} else {
currentNode.next = currentNode.next.next;
}
} else {
currentNode = currentNode.next;
}
}
return;
}
The problem here is with
if (!duplicate) {
duplicate = true;
currentNode = currentNode.next;
}
you are marking duplicate = true and immediately assigning the "currentNode = currentNode.next;"
due to this reference is getting preserve of the next node
So
1. Put the condition outside of the loop to check whether the head element itself is
that node, if->yes mark isDuplicate = true and proceed in the loop.
2. Inside the loop check afterward and then assign the next node.
Hope this should work
You skipped head node. Try replace
Node currentNode = head;
with
Node currentNode = new Node();
currentNode.next = head;
You should update the current node reference as well as head->next should point to the current node after removing the node.
try the below code:
if (!duplicate) {
duplicate = true;
currentNode = currentNode.next;
head.next= currentNode.next;
}else {
currentNode.next = currentNode.next.next;
currentNode = currentNode.next;
head.next = currentNode; }
`
I'm working on a project that requires me to implement merge-sort on a linked list and I am using the code from this post Here to help me. Can someone explain to why on line 6, when I call return merge(merge_sort(head),merge_sort(sHalf)); the method merge_sort(head) inside of it, which contains the same head pointer doesn't cause an infinite loop? It seems to me that it is starting all over again with the same head pointer.
public Node merge_sort(Node head) {
if(head == null || head.next == null) { return head; }
Node middle = getMiddle(head); //get the middle of the list
Node sHalf = middle.next; middle.next = null; //split the list into two halfs
return merge(merge_sort(head),merge_sort(sHalf)); //recurse on that
}
//Merge subroutine to merge two sorted lists
public Node merge(Node a, Node b) {
Node dummyHead, curr; dummyHead = new Node(); curr = dummyHead;
while(a !=null && b!= null) {
if(a.info <= b.info) { curr.next = a; a = a.next; }
else { curr.next = b; b = b.next; }
curr = curr.next;
}
curr.next = (a == null) ? b : a;
return dummyHead.next;
}
//Finding the middle element of the list for splitting
public Node getMiddle(Node head) {
if(head == null) { return head; }
Node slow, fast; slow = fast = head;
while(fast.next != null && fast.next.next != null) {
slow = slow.next; fast = fast.next.next;
}
return slow;
}
It's because of the previous line:
Node sHalf = middle.next; middle.next = null;
Specifically, the middle.next = null; part.
Understand that even though the head pointer is the same, we are splitting the list into half, using middle.next = null. So, in the next recursive call, it's only half the linked list which was sent originally.
And at one point, it will reach head.next == null condition.