I have written a program that removes a node in the single linkedlist given the node.
public class Solution {
/**
* #param node: the node in the list should be deleted
* #return: nothing
*/
public void deleteNode(ListNode node) {
// write your code here
// if node.next==null, we cannot delete the current node without given the previous node
if(node == null || node.next == null) return;
ListNode next = node.next;
node.val = next.val;
node.next = next.next;
// I wonder if this link needs to be removed as well
next.next = null;
}
}
The problem is pretty simple. However, many code samples online do not contain this line I wrote:
next.next = null;
Without this line, we already remove the node. After that, although nothing is pointing to "next", "next" still points to next.next. Without setting next.next = null, will the Java garbage collector remove this deleted node?
Indeed it will. The gc iterates over all objects and checks wether someone else points to it. If not, it is marked for deletion.
Related
Below is a program to reverse a doubly linked list in gfg.
When I try to submit, I get the following error
Your program took more time than expected. Time Limit Exceeded
Expected Time Limit 0.00sec.
Can anyone tell me how to decrease my program's time?
Code
public static Node reverseDLL(Node head)
{
Node p,cur;
p=head;
cur=head.next;
while(cur!=null)
{
if(cur.next==null)
{
p.next=cur.next;
cur.next=head;
head.prev=cur;
head=cur;
break;
}
p.next=cur.next;
cur.next.prev=p;
cur.next=head;
head.prev=cur;
head=cur;
cur=p.next;
}
return head;
}
Your code is not correct:
The prev member of the (original) tail node is not set correctly. It is not set in the if block, but in the previous iteration, where it got its value with cur.next.prev=p, which is setting it to the (original) head node. This creates an infinite loop in the data structure. In the end, none of the prev links is null. It might be that the testing framework keeps following those prev links in circles until a time out happens.
Also, the function assumes that head is not null. This may not be guaranteed.
There are also too many assignments happening. With a doubly linked list it is quite simple: just swap the value of next and prev in every node:
public static Node reverseDLL(Node head)
{
if (head == null) {
return null;
}
Node next = head.next;
while (next != null) {
head.next = head.prev;
head.prev = next;
head = next;
next = head.next;
}
head.next = head.prev;
head.prev = null;
return head;
}
this is classic solution for O(N) time, so I guess time constraint for this task is required to. solve it for O(1).
Take a look for this thread: https://www.google.com/url?sa=t&source=web&rct=j&url=https://discuss.codechef.com/t/reverse-a-doubly-linked-list-in-o-1/72850&ved=2ahUKEwjipdrQw_T1AhVYLTQIHahIDx8QFnoECBQQAQ&usg=AOvVaw1c0BDUotM0suEK7I4B9pQs
This question also applies to various linked lists methods as well. So, when I have a method:
public void insert(String key)
{
if(top == null)
{
top = new Node(key);
}else {
Node newNode = new Node(key);
Node rover = top;
Node prev = top;
boolean wentLeft = true;
while(rover != null)
{
if (rover.getName().compareTo(key) < 0)
{
prev = rover;
rover = rover.getRight();
wentLeft = false;
}else {
wentLeft = true;
prev = rover;
rover = rover.getLeft();
}
}
if(wentLeft == true)
{
prev.setLeft(newNode);
}else {
prev.setRight(newNode);
}
}
nElems++;
}
How is the top of the Binary Search Tree and its children updated despite not being directly set anywhere in the method?
I know this likely has something to do with shallow copying and that like rover/prev are still referencing top in memory but I still don't really get it.
Even though I feel like I understand Linked Lists and Binary Search Trees on a conceptual level I don't feel comfortable moving forward with them without understanding this.
There are no copies being made. When you assign prev = top, that only creates another reference to the same object as top, not a copy.
The code works because nodes are inserted one by one.
When prev.setLeft/setRight is called, prev is already in the tree because it was inserted before. So prev is already in the tree, i. e. the parent of prev is top, or the parent of the parent of prev, you get the idea. Thus, when new_node becomes child of prev, it becomes part of the tree.
This is what makes linked lists and trees so useful. When you insert an element, you only have to make one connection.
I am working on a problem: create a stack using a linked chain that places the new entry (top of stack) at the end of the chain, not the beginning. Part of the solution involves maintaining a reference to the first and last nodes and the next to last node. Also, I cannot traverse the chain. I cannot think of a solution because I cannot think of a way to update the reference to the next to last node when the top entry is popped off the stack?
(This is a textbook question. The beginning of the chain contains a reference to the next node. The end of the chain's reference to the next node is null)
Beginning and end are abstract terms. Just initialize a node to contain the first element. Something like the following:
Node tail = null;
..
..
Then to add a new node.
public void push(T val) {
Node = new Node(val);
Node.next = tail;
tail = node;
}
You're adding it at the tail and then the new node becomes the tail. Then when you iterate thru it you simply start at the tail and move towards the head using the next reference.
To pop the stack.
public T pop() {
if (tail != null) {
object = tail.object;
tail = tail.next;
}
return null;
}
You will need to create a generic class Node. And you may want to have an isEmpty() method.
Example input: the node ‘c’ from the linked list a->b->c->d->e Result: nothing is returned, but the new linked list looks like a->b->d->e
I do understand that ppl have already asked this question before, but since my reputation is not high enough yet, I couldn't ask my question in that thread. so here goes my quesetion:
So in the solution, when deleting the middle node we do:
public static boolean deleteNode(LinkedListNode n) {
if (n == null || n.next == null) {
return false; // Failure
}
LinkedListNode next = n.next;
n.data = next.data;
n.next = next.next;
return true;
}
But what I don't understand is that why can't I just do n = next?
It is probably a trivial question, but I didn't seem to find a good explanation for this question
If you just do n = next then you have only changed what object your local reference variable n refers to; you haven't modified any part of the list.
The trick to "deleting" the current node is to overwrite it with the next one:
n.data = next.data;
n.next = next.next;
Now you are modifying fields of the object that is referred to by n, which is a part of the actual list.
In C++, the code you wrote would look like this:
bool deleteNode(LinkedListNode* n) {
if (n == null || (*n).next == null) {
return false; // Failure
}
LinkedListNode* next = (*n).next;
(*n).data = (*next).data;
(*n).next = (*next).next;
return true;
}
So what does that mean? When you call this method, in C++ it would look like this:
LinkedListNode* listNode = new LinkedListNode();
deleteNode(&listNode);
This is important, because that means you're just sending an address over, and not the entire object. This means that you don't actually have access to the node you gave as a parameter to the method, you only have a reference to its address.
Basically, in Java, you can't do the following C++ code:
*n = *next;
You can't modify the listNode object that's outside of the method. You only get its address. And you are only modifying the copy of its address, not the address itself.
Basically, it's because in Java, the pointer of the class is passed by value (as a copy), and primitives are also passed by value (as a copy).
somehow, it overwrites the current node which is supposed to be deleted with the
data of next node to it,and delete the next node.
LinkedListNode next = n.next;
n.data = next.data;
n.next = next.next;
That is how the code comes.
I am attempting to remove the last iterated element, with a custom iterator/linked list class. It for some reason only does this for the first item in the list (the head condition). Is there anything wrong with the conditions?
Should I, instead of the Else after If (prev=head), write If (next != null) to find middle nodes, and If (next = null) to find the last node?
Second question: to remove the items, should I also write prev.element = null (now I only have prev = null, and I suppose that erases the node but not its content.
Quite simply, what is wrong with my remove method, as I cannot figure it out myself. Thank you guys so much in advance. I have been working many hours with this but I still haven't got it working.
public E next() {
if (!hasNext())
throw new NoSuchElementException ();
prev = next;
E element = next.element;
next = next.next;
return element;
}
public void remove() {
if(prev == null) {
throw new IllegalStateException();
}
else {
if(prev == head){
head = head.next;
next = head;
}
else {
next = prev.next;
}
sizeOfList--;
prev = null;
}
}
You would need a while loop to be able to go through every node in the list until you hit the last one. As it is now, your code simply goes past the head, and then gets into the code that says sizeOfList-- and then prev = null;
You need something like this:
while (prev.next.next != null) {
prev = prev.next;
}
prev.next = null;
I do prev.next.next so that you can set the 2nd to last node in your linked list to point to a null value (which is done by prev.next = null;). Think of it this way: prev is the 2nd to last element in the list, prev.next is the last element, and obviously prev.next.next HAS to be null (because prev.next is LAST.) So once this is the case, delete the last element by setting the 2nd to last element to point to a null value.
And then decrement your list count.
This is my best guess with the given code
if(prev == head){ should change to if(prev.equals(head)){ Use equals method.
And I think you have to override equals method in the corresponding element class might definitely help.
== only checks for whether both variables refer to same object in memory, where as equals check Object state.
I hope it helps :).