Rearrange a linked list such that all even and odd positioned nodes are together

Problem Statement

In this problem, we will be given a linked list, and we need to rearrange the nodes such that the nodes present at odd positions and the nodes present at even positions are together.

Problem Statement Understanding

To understand this problem statement, let us take examples.

If the given linked list is:

then according to the problem statement:

  • Starting the counting from 1, the nodes at odd positions are 3,18, and 5.
  • Nodes at even positions are 1 and 12.
  • So, after rearrangement, the resultant list will be:

Let us take another example:
If the linked list is 4→1→10→42→5→NULL.

  • Starting the counting from 1, the nodes at odd positions are 4,10, and 5.
  • Nodes at even positions are 1 and 42.
  • So, after rearrangement, the resultant list will be 4→10→5→1→42→NULL.

Now, I think the problem statement is clear, so let's see how we can approach it. Any ideas?

  • If not, it's okay; we will see in the next section how we can approach it.

Let’s move to the approach section.

Approach

  • We will use two pointers, where at first, we will initialize these pointers with the address of the first and the second node of the linked list.
  • Now, we will iterate the list from the first node to the last node.
    • While iterating, we will connect the node pointed by each pointer to the node next to the adjacent node in its right.
    • This will ensure that all odd and even positioned nodes are with each other, respectively.
  • At last, we need to connect the tail of the odd list with the head of the even list.
  • Finally, after all the above steps, we will have our rearranged linked list with all the even and the odd positioned nodes together.

The approach is discussed in more depth in the algorithm section.

Algorithm

1) We will return NULL if the head is NULL, i.e., the input list is empty.
2) Initialize two pointers odd and even with the first and the second node of the list, respectively.
3) Initialize a pointer evenHead with the second node.
4) Run an infinite while loop and inside it:

  • If any one of odd, even, or even→next is NULL (i.e., we have reached the end of the list, so we will connect the last node of odd list to the first node of the even list), then attach the tail of odd to the head of even and break from the loop.
  • Connect odd to the node next to even and update odd with this node.
  • If the node next to odd is NULL (i.e., no node after the current odd node), then update the next of even as NULL and attach the tail of odd to the head of even and break from the loop.
  • Connect even to the node next to odd and update even with this node.
    5) Return head at last from the function.

Dry Run


Code Implementation

#include
using namespace std;
class Node{
    public:
    int data;
    Node* next;
    Node(int x){
        data = x;
        next = NULL;
    }
};

// Using this function we will print the linked list
void printList(Node *head)
{
    if (!head)
        return;
     
    while (head->next != NULL)
    {
        cout << head->data << ",";
        head = head->next;
    }
    cout << head->data << endl;
}

// Function to rearrange the linked list
Node *rearrangeEvenOdd(Node *head)
{
    // Empty list condition
    if (head == NULL)
        return NULL;
 
    Node *odd = head;
    Node *even = head->next;
 
    // first node of even list
    Node *evenHead = even;
 
    while (1)
    {
        // If we have reached end of the list
        // then, connect last node of odd list
        // to the first node of even list
        if (!odd || !even || !(even->next))
        {
            odd->next = evenHead;
            break;
        }
 
        odd->next = even->next;
        odd = even->next;
 
        // No nodes after current odd node
        if (odd->next == NULL)
        {
            even->next = NULL;
            odd->next = evenHead;
            break;
        }
 
        even->next = odd->next;
        even = odd->next;
    }
 
    return head;
}

int main(void){
    Node* head = NULL;
    head = new Node(3);
    head->next = new Node(1);
    head->next->next = new Node(18);
    head->next->next->next = new Node(12);
    head->next->next->next->next = new Node(5);
    cout<<"Original linked list without rearrangement: "<

						 

#include
#include

struct Node
{
    int data;
    struct Node* next;
};
 
// A utility function to create a new node
struct Node* newNode(int key)
{
    struct Node *temp = (struct Node *) malloc(sizeof(struct Node));
    temp->data = key;
    temp->next = NULL;
    return temp;
}
 
// Rearranges given linked list such that all even
// positioned nodes are before odd positioned.
// Returns new head of linked List.
struct Node *rearrangeEvenOdd(struct Node *head)
{
    // Corner case
    if (head == NULL)
        return NULL;
 
    // Initialize first nodes of even and
    // odd lists
    struct Node *odd = head;
    struct Node *even = head->next;
 
    // Remember the first node of even list so
    // that we can connect the even list at the
    // end of odd list.
    struct Node *evenFirst = even;
 
    while (1)
    {
        // If there are no more nodes, then connect
        // first node of even list to the last node
        // of odd list
        if (!odd || !even || !(even->next))
        {
            odd->next = evenFirst;
            break;
        }
 
        // Connecting odd nodes
        odd->next = even->next;
        odd = even->next;
 
        // If there are NO more even nodes after
        // current odd.
        if (odd->next == NULL)
        {
            even->next = NULL;
            odd->next = evenFirst;
            break;
        }
 
        // Connecting even nodes
        even->next = odd->next;
        even = odd->next;
    }
 
    return head;
}
 
// A utility function to print a linked list
void printlist(struct Node * node)
{
    while (node != NULL)
    {
        printf("%d->",node->data);
        node = node->next;
    }
    printf("NULL\n");
}
 
// Driver code
int main(void)
{
    struct Node *head = newNode(6);
    head->next = newNode(7);
    head->next->next = newNode(8);
    head->next->next->next = newNode(9);
    head->next->next->next->next = newNode(10);
 
    printf("Given Linked List\n");
    printlist(head);
 
    head = rearrangeEvenOdd(head);
 
    printf("\nModified Linked List\n");
    printlist(head);
 
    return 0;
}
class Rearrange
{
    static class Node
    {
	    int data;
	    Node next;
    }
    static Node newNode(int key)
    {
	    Node temp = new Node();
	    temp.data = key;
	    temp.next = null;
	    return temp;
    }
    /* Rearranges given linked list such that all even positioned
    nodes are before odd positioned returns new head of linked List.*/
    static Node rearrangeEvenOdd(Node head)
    {
	    // Corner case
	    if (head == null)
		    return null;

	    // Initialize first nodes of even and odd lists
	    Node odd = head;
	    Node even = head.next;

	    /* Remember the first node of even list so
	    that we can connect the even list at the end of odd list.*/
	    Node evenFirst = even;

	    while (1 == 1)
	    {
		    /* If there are no more nodes,
		    then connect first node of even
		    list to the last node of odd list*/
		    if (odd == null || even == null ||(even.next) == null)
		    {
			    odd.next = evenFirst;
			    break;
		    }
            // Connecting odd nodes
		    odd.next = even.next;
		    odd = even.next;
            // If there are NO more even nodes after current odd.
		    if (odd.next == null)
		    {
			    even.next = null;
			    odd.next = evenFirst;
			    break;
		    }
            // Connecting even nodes
		    even.next = odd.next;
		    even = odd.next;
	    }
	    return head;
    }
    static void printlist(Node node)
    {
	    while (node != null)
	    {
	    	System.out.print(node.data + "->");
		    node = node.next;
	    }
	    System.out.println("NULL") ;
    }
    // Driver code
    public static void main(String[] args)
    {
	    Node head = newNode(1);
	    head.next = newNode(2);
	    head.next.next = newNode(3);
	    head.next.next.next = newNode(4);
	    head.next.next.next.next = newNode(5);

	    System.out.println("Given Linked List");
	    printlist(head);

	    head = rearrangeEvenOdd(head);

	    System.out.println("Modified Linked List");
	    printlist(head);
    }
}




class Node:
	def __init__(self, d):
		self.data = d
		self.next = None

class LinkedList:
	def __init__(self):
		self.head = None
		
	def newNode(self, key):
		temp = Node(key)
		self.next = None
		return temp

	def rearrangeEvenOdd(self, head):
		
		if (self.head == None):
			return None

		odd = self.head
		even = self.head.next

		evenFirst = even

		while (1 == 1):
			
			if (odd == None or even == None or
							(even.next) == None):
				odd.next = evenFirst
				break

			odd.next = even.next
			odd = even.next

			if (odd.next == None):
				even.next = None
				odd.next = evenFirst
				break

			even.next = odd.next
			even = odd.next
		return head

	def printlist(self, node):
		while (node != None):
			print(node.data, end = "")
			print("->", end = "")
			node = node.next
		print ("NULL")
		
	def push(self, new_data):
		new_node = Node(new_data)
		new_node.next = self.head
		self.head = new_node

ll = LinkedList()
ll.push(5)
ll.push(12)
ll.push(18)
ll.push(1)
ll.push(3)
print ("Given Linked List")
ll.printlist(ll.head)

start = ll.rearrangeEvenOdd(ll.head)

print ("Modified Linked List")
ll.printlist(start)

Output

Original linked list without rearrangement:
3,1,18,12,5
Linked list after rearrangement:
3,18,5,1,12

Time Complexity: O(n), where n is the total number of nodes in the list
[forminator_quiz id="5828"]

So, in this blog, we have tried to explain how we can rearrange a linked list such that all even and odd positioned nodes are together most optimally. If you want to solve more questions on Linked List, which are curated by our expert mentors at PrepBytes, you can follow this link Linked List.

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