I am writing a simple recursive code to reverse a linked-list in Java. Strangely it returns the same node.
Node reverse(Node ptr) {
if (ptr.link == null)
return ptr;
Node second = ptr.link;
ptr.link = null;
Node rest = reverse(second);
rest.link = ptr;
return rest;
}
Why isn't this working?
Your current approach won't work, the rest value that's being returned from the recursive call doesn't point to the last element in the list (which should be the insertion point for the nodes), instead it's pointing to the first element of the result. In other words, this line isn't correctly building the output:
rest.link = ptr;
For successfully reversing a list, one strategy is to use an accumulator parameter, so we can insert at the head of the list as the recursion unwinds - bear in mind that the recursion starts to "go back" in reverse order with respect to the elements in the list. Try this, it's a tail-recursive solution:
Node reverse(Node ptr, Node acc) {
if (ptr == null)
return acc;
Node second = ptr.link;
ptr.link = acc;
return reverse(second, ptr);
}
Call the above method like this:
Node reversedList = reverse(list, null);
Be aware that the input list will be modified in-place and won't be the same after the method returns, any variable that holds a reference to list will be affected. If that's a problem, you should create a copy of list before passing it to reverse().
Oscar is correct that the return value from recursive call does not point to the last element.
There is another algorithm that does not use accumulator parameter:
function reverse(node) {
if (node.next == null)
return node;
var tail = reverse(node.next);
node.next.next = node;
node.next = null;
return tail;
}
Node reverseNode(Node node){
Node head = null;
While(node.next != null){
Node n = new Node(node.data);
n.next = head;
head = n;
node = node.next;
}
return head;}
You know you don't have to write a recursive function
Related
I am trying to sort my linked list with merge sort. The list is actually sorted but it is kind of missing first item(s).
Merge sort functions:
public Node mergeSort(Node head) {
if (head == null || head.next == null) {
return head;
}
Node middle = middleElement(head);
Node nextofMiddle = middle.next;
middle.next = null;
return merge(mergeSort(head), mergeSort(nextofMiddle));
}
public Node merge(Node left, Node right) {
Node temp = new Node();
Node newHead = temp;
while (left != null && right != null) {
if (left.info <= right.info) {
temp.next = left;
temp = left;
left = temp.next;
} else {
temp.next = right;
temp = right;
right = temp.next;
}
}
temp.next = (left == null) ? right : left;
return newHead;
}
public Node middleElement(Node head) {
if (head == null) {
return head;
}
Node slow = head;
Node fast = head;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
So I printed the list to the screen using traverse:
public static void main(String[] args) {
MyLinkedList mll = new MyLinkedList();
mll.insert(3);
mll.insert(5);
mll.insert(9);
mll.insert(1);
mll.insert(8);
mll.insert(7);
mll.insert(2);
mll.mergeSort(mll.head);
mll.traverse();
}
I have result like this:
1 and 2 missing!
After checking, i noticed that the "tail" of the linked list value is still 2. I don't know why can someone help?. I'm really new to programming so sorry for any inconvenience. Thank you for reading!
You're not far from a correct solution. I was able to get a working sort by adding 1 line and changing 3.
Your merge isn't doing what you think it is.
Allocating a false head node (what you call temp, not a good choice of name; try falseHead) is a fine idea. Then falseHead.next is the true head. That's what you'll finally return as the sorted list. Note that initially it's null, which is what you'd expect.
What you're missing is a variable tail to reference the current last node in the merged list. Since there is initially no last node at all, this should be initialized equal to falseHead. Consequently, when you append the first element to the tail of the current result by setting tail.next, you'll also be setting falseHead.next, i.e. creating the head element. It all works out.
Now, what is the logic for, say, removing the current head of the right list and appending it to the merge result? It's just 3 steps:
Do the append operation by making tail.next the current head of right.
Update tail to tail.next.
Remove the head of right by updating right to right.next.
Of course the left side is similar.
Good luck. You're close.
You have a couple problems that I can see. One, mentioned by #rcgldr, in the comments is that you merge method is returning the temp node newHead, but that node was no value and is not part of your graph. You need to return it's child newHead.next.
The other problem is that you never reset the head of your list. So after all the sorting, the head of mll still points to the original head, which in this case is 3. So when you traverse the list you skip 1 and 2. I'm not sure what your linked list class looks like but something like this should be close -- just assign the final node returned by mergeSort to the head of the list. Then when you traverse you should be starting in the right spot:
mll.head = mll.mergeSort(mll.head);
mll.traverse();
The problem is to reverse a singly linked list, but I'm a bit stuck with the following code:
public ListNode reverseList(ListNode head) {
if(head == null || head.next == null) return head;
ListNode temp = head.next;
head.next = null;
ListNode result = reverseList(temp);
temp.next = head;
return result;
}
What does line 6 do?
temp.next = head;
Without it, the result will only return the last element and not a reversed list.
The way I read it is that the code gets to line 5, gets result, goes through line 6, then returns the result at line 7, and keeps repeating until finish. How does line 6 actually affect the code?
I'm having a hard time visualizing this recursion. Can someone help?
A recursive function solves a problem by making use of smaller versions of the exact same problem. There are two main parts to a recursive function: the base case and the recursive case. The base case is the smallest version of the problem where we provide a solution without having to use recursion. It's where the recursion stops. The recursive case makes the assumption that the smaller versions of the problem provide correct solutions and we use those solutions to solve the larger versions.
reverseList() takes the head node of a linked list as input, reverses each node in the list, and returns the head node of the new reversed list.
Let's say you have a linked list:
a->b->c->d->e
head points to node a and temp points to node b.
On line 4, node a is disconnected from the original list resulting in two sublists, one containing only a and the other containing the other elements starting with b. Now, assuming that line 5 correctly reverses the sublist starting with b, we have:
a
e->d->c->b
head still points to a, temp still points to b, and result points to e (the head of the reversed sublist).
In order to correctly reverse our original list, we just add node a (head) to the end of the reversed sublist (temp).
Now the list is:
e->d->c->b->a
where the new head points to e.
Line 6 only make link between the nodes. recursion goes all the way to last node and make link reverse.
EDIT : Posted full code.
I wrote an alternative that you may understand better. I also set the function as static ( getters and setters not used for more straightforward explanation).
public class ListNode {
// should be private variables
int value;
ListNode next;
ListNode(int val, ListNode nodeList){
this.value = val;
this.next = nodeList;
}
public static ListNode reverseList(ListNode head) {
if( head.next.next != null)
reverseList (head.next);
head.next.next = head;
head.next = null;
return head;
}
#Override
public String toString(){
String ret = "";
if ( next != null ){
ret = next.toString();
}
ret += String.valueOf(value);
return ret;
}
}
public class Main {
public static void main(String[] args) {
ListNode n0 = new ListNode(4, null),
n1 = new ListNode(3, n0),
n2 = new ListNode(2, n1),
n3 = new ListNode(1, n2),
n4 = new ListNode(0, n3);
System.out.println(n4);
ListNode.reverseList(n4);
System.out.println(n0); // <-- n4 now is the last item, check n0 instead !
}
}
Output
43210
01234
It may be easier for you to understand it here. Drawing it the before and after always helps. Like this, but only changing the arrows insted of the position of the boxes.
I tried to implement a circularly linked list and I had read that the complexity of removing the tail (which has a direct reference) is O(1). However I can't get rid of the tail reference without looping and resulting in a complexity of O(n) and so I was wondering if this is at all possible without looping?
remove = tail;
prev = temp = tail.getNext();
do
{
prev = temp;
temp = temp.getNext();
}
while(!temp.getNext().equals(tail.getNext()));
prev.setNext(temp.getNext());
size--;
return remove;
Note: prev, temp, removeare local and tail is the tail of the list and first instance of tail.getNext() is the head.
I had tried the below as well but it does not work as the node before tail still points to tail and the reference isn't removed. I know that I need make the tail's previous point to the head in order to remove the tail and the above code does this, but I'm just unsure how to do this and if it's possible without looping through the list..
remove = tail;
tail= lastNode.getNext();
size--;
return remove;
In a circularly linked list every node has a previous and next. There is the head pointer to the first node, and its previous is the tail node, whose next is the head node.
So
Node removeTail() {
if (head == null) {
return null;
}
Node tail = head.previous;
Node newTail = tail.previous;
newTail.next = head;
head.previous = newTail;
if (head == tail) {
head = null;
}
--size;
return tail; // If so desired.
}
I've a linked list that has two links - next and another assume. Initially, all another hold null. Now, I'm trying to convert this to a binary tree with next holding the liftChild and another holding rightChild. However, I want to do this in O(n) and in constant space. I've tried a lot. Below is my code. I'm verifying the result by level-order traversing the resulting tree. Currently, I know what's the mistake but don't know how to solve it. The mistake here is that inside the while loop, I'm changing the links of the node properly, but this means I cannot do node = node.next at the end because node's next is already pointing somewhere ahead in the list. So, I don't know how to traverse every node. Any hint, help is appreciated. Not hw, not interview question or anything. Just trying to learn data structures. So, this tree and ll stuff!
public class LlToBt {
public SpecialNode llToBt(SpecialNode node) {
SpecialNode temp = node;
SpecialNode returnNode = node;
if(node == null)
return null;
if(node.next == null)
return node;
node = returnNode;
while(node != null) {
SpecialNode currentNode = node;
temp = temp.next;
if(temp == null) {
//node.next = null;
//node.another = null;
return returnNode;
}
node.next = temp;
temp = temp.next;
if(temp == null) {
//node.next = null;
//node.another = null;
return returnNode;
}
node.another = temp;
node = currentNode.next;
}
return returnNode;
}
}
If you are changing the link that you will want to traverse before you traverse it, then you need to store off in a different variable the result of the traversal before you change the link. Then after the change, you can traverse using your temporary copy.
Given the current node, how can I find its previous node in a Singly Linked List. Thanks. Logic will do , code is appreciated. We all know given a root node one can do a sequential traverse , I want to know if there is a smarter way that avoids sequential access overhead. (assume there is no access to root node) Thanks.
You can't.
Singly-linked lists by definition only link each node to its successor, not predecessor. There is no information about the predecessor; not even information about whether it exists at all (your node could be the head of the list).
You could use a doubly-linked list.
You could try to rearrange everything so you have the predecessor passed in as a parameter in the first place.
You could scan the entire heap looking for a record that looks like a predecessor node with a pointer to your node. (Not a serious suggestion.)
If you want to delete the current node, you can do that without finding previous node as well.
Python Code:
def deleteNode(self, node):
node.val = node.next.val
node.next = node.next.next
# Delete Node in a Linked List
Walk through the list from the beginning until you meet a node whose next link points you your current node.
But if you need to do this, perhaps you oughtn't be using a singly linked list in the first place.
Your only option for a singly-linked list is a linear search, something like below (Python-like pseudo code):
find_previous_node(list, node):
current_node = list.first
while(current_node.next != null):
if(current_node.next == node):
return current_node
else:
current_node = current_node.next
return null
Assuming that you're talking about a forward singly linked list (each node only has a pointer to 'next' etc) you will have to iterate from the start of the list until you find the node that has 'next' equal to your current node. Obviously, this is a slow O(n) operation.
Hope this helps.
Keep two-pointer(curr, prev) initially both will point to head of the list.
do a loop on the list until you either reach at the end of the list or at the required node.
for each iteration move curr node to the next of it but before moving to next store its pointer in prev pointer.
prev = curr; // first store current node in prev
curr = curr->next // move current to next
at the end of loop prev node will contain previous node.
getPrev(head, key) {
Node* curr = head;
Node* prev = head;
while(curr !=null && curr->data==key){
prev = curr;
curr = curr->next
}
return prev
}
Example:
list = 1->5->10->4->3
We want prev node of 4 So key = 4 and head point at 1 here
initially:
temp will point to 1
prev will point to 1
iteration 1:
First, assign prev=temp (prev point to 1)
move temp; temp->next (temp point to 5)
iteration 2:
First, assign prev=temp (prev point to 5)
move temp; temp->next (temp point to 10)
iteration 3:
First, assign prev=temp (prev point to 10)
move temp; temp->next (temp point to 4)
iteration 4:
temp->data == key so it will return out of loop.
return prev node
This is some kind of hack which I found out while solving the problem(Delete every even node in a list)
internal void DeleteNode(int p)
{
DeleteNode(head, p);
}
private void DeleteNode(Node head, int p)
{
Node current = head;
Node prev = null;
while (current.next != null)
{
prev = current;
current = current.next;
if (current.data == p)
{
prev.next = current.next;
}
}
}
Now here, in prev you assign the current and then move the current to next thereby prev contains the previous node.
Hope this helps...
You can do it like this.. you can replace the value of current node by value of next node.. and in the next of 2nd last node you can put null. its like delete a element from a string. here is code
void deleteNode(ListNode* node) {
ListNode *pre=node;
while(node->next)
{
node->val=node->next->val;
pre=node;
node=node->next;
}
pre->next=NULL;
}
Use a nodeAt() method and pass the head,size and the index of the current node.
public static Node nodeAt(Node head,int index){
Node n=head;
for(int i=0;i<index;i++,n=n.next)
;
return n;
}
where n returns the node of the predecessor.
Here is a small trick with linear search: just pass in the node or its position whose previous node you are searching for:
private MyNode findNode(int pos) {
//node will have pos=pos-1
pos-- = 1;
MyNode prevNode = null;
int count = 0;
MyNode p = first.next; // first = head node, find it however you want.
//this is for circular linked list, you can use first!=last for singly linked list
while (p != first) {
if (count == pos) {
prevNode = p;
break;
}
p = p.next;
count++;
}
return prevNode;
}
We can traverse through the LinkedList using slow and fast pointers.
Let's say
fast pointer fast = fast.next.next
slow pointer slow = slow.next
Slow pointer will be always a previous of the fast pointer, and so we can able to find it.
It can possible to deleteNode if only given node not root or head. How ?
It can achieve by reversing value in node
4-> 2 -> 1 -> 9 given 2 as node to remove. as above other can't access previous node which is correct because singly linked list we don't store predecessor. What can do is swap value of next of give node to given node and change link to next of next of give node
nextNode = node.next // assigning given node next to new pointer
node.val = nextNode.val // replacing value of given node to nextNode value
node.next = nextNode.next // changing link of given node to next to next node.
I tried this approach and its working fine.
assuming you are using forward singly linked list your code should look like
while(node)
{
previous = node
node = node.next
// Do what ever you want to do with the nodes
}