I am new to programming, so can anyone explain me how the below code works?
Node reverse(Node node) {
Node prev = null;
Node current = node;
Node next = null;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
node = prev;
return node;
}
I just want to know exactly what is happening in the while loop.
Thanks in advance.
Basically it tries to reverse a linked list and as you probably know, every Node has two pointers: a next and a prev. In order to reverse the linked list, we should change these pointers in a way that you can see in the below picture(we want to change blue chain to red chain)
So for example imagine Node 3 in the picture, what its prev field has been pointing to (Node 2) is going to be the next node of Node 3 (in our desired red chain), so current.next = prev;(line 2) makes sense. But because we want to get to Node 4 in the next iteration of the loop, before that we write next = current.next;(line 1) to save the next pointer of Node 3 which currently points to Node 4, and finally at the end of each iteration we put back that with current = next;(line 4) to iterate to the next Node.
In the next iteration we need to do the same thing so now prev should point to Node 3 and that's why we had prev = current;(line 3) to prepare prev.
Picture reference
For each node in the list it is changing the next pointer to point to the previous node.
eg. It is taking this linked list:
a->b->c
and changing it into:
a<-b<-c
Related
For the first function, shouldn't "prev = head" be outside of else because we want to set the previous every time before we change the head value?
For the second function, shouldn't "p2 = p2.next" be outside of else because we want to go next every time?
Thank you guys.
//This would take O(n) but would require extra space O(n)
public static Node removeDuplicates(Node head){
Node prev = null;
Set<Integer> hs = new HashSet<Integer>();
while(head!= null){
if(hs.contains(head.data)){
prev.next = head.next;
}
else{
hs.add(head.data);
//why is prev = head here instead of out of the else statement?
prev = head;
}
head = head.next;
}
return head;
}
//This would take O(n^2) but no extra space is required.
public static Node removeDuplicatesTradeOff(Node head){
//pointer 1 and pointer 2.
Node p1 = head;
while(p1.next != null){
Node p2 = p1;
while(p2.next != null){
if(p1.data == p2.next.data){
p2.next = p2.next.next;
}
else{
//why is p2 = p2.next here instead of out of the else statement?
p2 = p2.next;
}
}
p1 = p1.next;
}
return head;
}
Shouldn't "p2 = p2.next" be outside of else because we want to go next every time
I think it would be more accurate to say that we want to have a different next available every time, instead of saying that we want to "go next" every time.
We don't always want to "go next". When prev.next has changed due to another operation (in this case removal of a duplicate) we want to stay where we are, because prev.next has already changed and is now pointing to a node further ahead (because a duplicate node has just been removed).
In other words, we don't want to have a different prev every time, we just want to have a different prev.next every time. So as long as prev.next advances every time, we don't care if prev stays the same sometimes.
That's why in both methods prev (or p2) only advances in the else branch, while prev.next (or p2.next) is updated (advances) only in the if branch.
Think of these two as different operations, the else branch being "go next" and the if branch being "drop next". When you drop a node ahead of you, you did not move (true!), but since you did drop one node ahead of you, now there is a new node ahead of you, so did not have to move. So you can just continue with if/else checks and sooner or later you will reach the end or you will drop the last node ahead of you.
One input example to illustrate this point is
head(1) -> 1 -> 1 -> 1 -> null
With such an input, the algorithm would only do
drop next
drop next
drop next
and it would be done.
No "go next" happened at all.
Result
head(1) -> null
Using solution 1 as a reference because the answer applies to both solutions. When you find a duplicate at the current node (head), you set the previous node’s (prev) next node to the current node’s next node (which removes the duplicate). On the next iteration, you will go to the next node in the list which is already being pointed at by the previous node. So, there is no need to overwrite prev. Another way to think about it is that the head you would be setting prev to is the node that you just removed. You wouldn’t want to do that.
Before removal:
node1 (prev) -> node2(head) -> node3
After removal:
node1 (prev) -> node3 (head)
As you can see, prev stays the same. No need to update it.
removeDuplicates:
If we remove a duplicate, the current removed node should not be assigned to prev. In fact prev is the previous node in the remaining kept list of uniques.
As is entirely evident, when there are more than 2 repeated values.
... -> prev: [42] -> [42] -> [42] -> [51] -> ...
should become
... -> prev: [42] -> [51] -> ...
removeDuplicatesTradeOff:
The same case. p2 (the previous node) may only advance when no duplicate was found.
Then there is a tiny bug:
removeDuplicates should return the new head normally - in case the head node is removed:
head = remove...(head); // Must assign, should the head node itself be removed.
Now it returns null. However for duplicates, the head node is never deleted. So either make it a void function or return the old head.
public static Node removeDuplicates(Node head){
Node prev = null;
Set<Integer> hs = new HashSet<>();
Node current = head;
// Loop-invariant: hs.isEmpty() || prev != null
while (current != null) {
if (hs.contains(current.data)) {
prev.next = current.next;
} else {
hs.add(current.data);
prev = current;
}
current = current.next;
}
return head;
}
Also notice that it is bad style using the name head for a running pointer.
I have the following method:
public static IntNode reverse (IntNode head) {
IntNode prev = null;
IntNode curr = head;
IntNode next = null;
while(curr!=null) {
next = curr.link;
curr.link = prev;
prev = curr;
curr = next;
}
return prev;
}
This works perfectly but when run twice it just returns the first (technically last) node in the linked list and nothing more, why is this the case and could you help me fix this? Thanks.
For Example :
Before reverse:
10
25
5
14
2
After reversal:
2
14
5
25
10
After typing reversal again:
10
I think the method is correct.Just make sure whenever you call the reverse() method, assign the result of reverse() method to its head.
head = reverse(head);//correct coz head changes after reversal
when you fail to assign the result of reverse() method to head.
reverse(head)//Wrong coz head won't change after reversal
the head will not be updated and it will point to tail of linked list only.Hope this helps.
Your entire list is starts off with head. Once you switch your nodes around, head is now pointing to the last node (i.e. its .link is null).
You need to reset head as the first node in order for you to access the list.
This question already has answers here:
What is a debugger and how can it help me diagnose problems?
(2 answers)
Closed 3 years ago.
I have a complete working method to reverse a doubly linked list. To be honest I've been going back and fourth for months trying to trace this code to see exactly how it works but I get confused at the end of the while look when I update my current node with current.prev
Ive tried to print out the values of the nodes for each time it changes the next and previous pointers however I get a nullpointerexception, so no luck there.
public void reverse(){
Node temp = null;
Node current = head;
while(current != null){
temp = current.prev;
current.prev = current.next;
current.next = temp;
current = current.prev;
}
if(temp != null){
head = temp.prev;
}
}
There are no errors here, I passed it thru my own test cases for the worst and best scenario. I just can't seem to understand what is going on. I know this is essentially swapping the next and prev pointers but I need to know how.
public void reverse(){
// Create initial values
Node temp = null;
// Note that you are using current to traverse through the linked list
Node current = head;
// While current is not at the end of the original (non-reversed) list
while(current != null){
// Swapping prev and next
temp = current.prev; // temp 'temporarily' holds copy of current.prev
current.prev = current.next; // current.prev is overwritten with current.next
current.next = temp; // current.next is overwritten with temp (containing original current.prev)
// You are setting current to the newly redefined prev
// This was equal to current->next in the original (non-reversed) list
// So you are traversing through the original list
// Anything 'before' this has already been reversed
// Anything 'after' still needs to be reversed
current = current.prev;
}
// Condition checks for edge case of a one node linked list
if(temp != null){
// Set the head of the reversed list
head = temp.prev;
}
Commented code above. I'm not a Java programmer, but in C I would print out the addresses of each node before and after the reversal to check I am doing things properly. Perhaps, you can use hashcodes to do something similar?
This is similar to a swap in Java:
while(current != null){
temp = current.prev; //gets the 'end' of the doubly linked list
current.prev = current.next; //sets the 'end' of the doubly linked list to the 'first' of the list
current.next = temp; //sets the 'first' of the list to temp, which is the 'end' of the list
current = current.prev; //iterate through the list in reverse order (similar to i++ in for loops)
}
if(temp != null){
head = temp.prev; //edge case when there is only 1 node.
}
Clarification: The main question is which node is something like current.previous.next = current.next actually pointing to?
I'm using some sample code I found on a YouTube video, but really trying to break it down and understand it. Everything works as-is, I've added comments to each section to help me understand. Where I am really running into problems explaining what is happening in plain English is for node removal when the code start using previous and next in the same line. I'm not following exactly what it's pointing to now. I've worked through the code in the debugger, but I need some plain English explanation if someone wouldn't mind.
Let's say for example I've got a DLL that has 3,4,5,6,7. So I'm going to remove 5, which is the 3rd index. Here is the method for removal.
//Method to remove node at a specified position
public void removeAt(int index) {
//If the head doesn't exist, break out of the logic
if(head == null) return;
//If the requested index is smaller than 1 or greater
//than the size of the list, break out.
if(index < 1 || index > size) return;
//Declares the currently used link as the head
Link current = head;
//Declare int i counter for use in while loop
//While i is less than the index set current to the next node
//and add to the counter i
int i = 1;
while(i < index) {
current = current.next;
i++;
}
//If the next node doesn't exist, set current...previous next?
if(current.next == null) {
current.previous.next = null;
}
//Else if the node before the current is null, set current to
//the next node and then set the previous to null (I thought it
//already was null??) Set the head to the current node
else if(current.previous == null) {
current = current.next;
current.previous = null;
head = current;
}
//If none of the above conditions, set current previous next?? to
//the next node and current next previous to the previous node??
else {
current.previous.next = current.next;
current.next.previous = current.previous;
}
//Subtract from the size of the list
size--;
}
My main understanding issues come in when it starts using current.previous.next and current.next.previous. To me, current.previous.next is just saying to remain at current. If I have three numbers, 3 4 5 and current is 4, then previous is 3, so next would just go back to 4. I know this isn't right, but after reading other posts here, the Javadoc, and watching videos, I'm still not understanding what is going on.
Here is the other class:
public class Link {
private int data;
public Link previous;
public Link next;
public Link(int data) {
previous = null;
this.data = data;
next = null;
}
public Link(Link previous, int data, Link next) {
this.previous = previous;
this.data = data;
this.next = next;
}
public int Data() {
return data;
}
}
I'd appreciate some explanation. Thanks!
If node.next is null, it means that the node is the last element of the list. So if you try to remove the last element of the list, the element before the last element becomes the last element. Which is what this code does
if(current.next == null) {
current.previous.next = null;
}
If node.previous is null, it means that it is the first element of the list. First, we have to keep a reference to the following element, define that the following element must not have a previous element, and then define it as the head of the list. Which is what this code does
if(current.previous == null) {
current = current.next;
current.previous = null;
head = current;
}
Then to stay on your example
if current is 4, then current.previous is 3 and current.next is 5
so
current.previous.next = current.next
define that 3.next is now 5
and
current.next.previous = current.previous
define the 5.previous is now 3
So then 4 is not reference anymore in the list
Doubly link list (DLL) is a list that can be traversed in both directions as you already know. Each node has previous and next pointers to their siblings allowing navigation to both directions. You already have head variable pointing to the head of the linked list.
In the remove method you passed in the index of the node you wish to delete.
while(i < index) {
current = current.next;
i++;
}
Above while loop sets the current node to be the node we wish to delete. There are three cases to be handled at this point.
Current node can be the first node in the list. So current.previous points to nothing and current.next may point to next node if there is one.
Current node can be the last node of the list so that current.next = null.
Current node is neither head or tail of the list so that current.next and current.previous have corresponding values.
if(current.next == null) {
current.previous.next = null;
}
Above code section will is the second point mentioned above. current points to last node. In order to remove last node you have to go back to current.previous and set its link to null so that current will no longer be accessible from the previous node.
else if(current.previous == null) {
current = current.next;
current.previous = null;
head = current;
}
checks to see if the current node is the head of the node. If current is pointing to head of the linked list, it does not have any previous sibling so that current.previous points to null. So current.next must become the new head of the list. With current.previous = null we set the previous link to null and set this current head as the head variable. Now previous node has gone. No references to it.
Third case is the above bullet point 3. It's handled by
else {
current.previous.next = current.next;
current.next.previous = current.previous;
}
where we want to remove current node. That means we need to link current.previous and current.next as siblings.
As a final note, your Link class uses public fields. They should be private and access via getters and setters.
I have to create a method that inserts information numerically but, the problem specifically says not to use a sorting method.
The list should always be sorted by rfidTagNumber. However, this doesn't mean you run a sorting algorithm on the list. As you insert each info into the sorted list, traverse the list to figure out where the new info should go and insert there. Then the new list is still sorted without running a sorting algorithm.
Also, I have a node that has multiple components. How can I make one node greater than the other based on one component such as price. (Price is one of the components)
I'm new to linked lists so I'm struggling here..
Thanks guys
if (head == null) // if the list is empty then you can just put the info in the first link
head = temp = cursor = null;
else { //other options
while (temp != null)
for (temp = head; temp.next != null; temp = temp.next );
if ( temp > temp.prev && temp < temp.next )
temp = temp.next;
}
^ this is what I have right now. Its full of errors but what I want to do is pretty much say that if a node is greater than the previous and less than the next one then, insert here.
How can I implement this and how can I make the RFDNumber the metric in which the computer decides whether or not a node is greater than another.
Just start at the beginning of the list and at each step compare your rfidTagNumber to number for the next node. If it is less than or equal to that node's number, insert the new node after the current node.
If you give each node a public rfidTagNumber member, you can reference it directly in your conditional. Also, your function should have two parameters: a head node head that refers to the start of the list, and a node you want to insert, let's call it insertNode.
Given these two values, let's change the conditional of your function a bit. You do not want to compare insertNode to the previous and next node, you want to compare it to the current and next node. So instead of:
if ( temp > temp.prev && temp < temp.next )
We'd do something like:
if ( insertNode.rfidTagNumber >= temp.rfidTagNumber && insertNode.rfidTagNumber <= temp.next.rfidTagNumber )
If this condition is true then you want to insert your node here. This involves changing next for the current node and prev for the next node. Here we also need to properly assign prev and next for insertNode. We can do this with the following code:
insertNode.prev = current
insertNode.next = current.next
current.next.prev = insertNode
current.next = insertNode
Keep in mind the order of these statements matters as you do not want to modify current.next without first storing a reference to it in insertNode.next.