Java Program For Rearranging A Linked List Such That All Even And Odd Positioned Nodes Are Together
Rearrange a linked list in such a way that all odd position nodes are together and all even positions node are together,
Examples:
Input: 1->2->3->4
Output: 1->3->2->4
Input: 10->22->30->43->56->70
Output: 10->30->56->22->43->70
The important thing in this question is to make sure that all below cases are handled
- Empty linked list.
- A linked list with only one node.
- A linked list with only two nodes.
- A linked list with an odd number of nodes.
- A linked list with an even number of nodes.
The below program maintains two pointers ‘odd’ and ‘even’ for current nodes at odd and even positions respectively. We also store the first node of the even linked list so that we can attach the even list at the end of the odd list after all odd and even nodes are connected together in two different lists.
Java
class GfG {
static class Node
{
int data;
Node next;
}
static Node newNode( int key)
{
Node temp = new Node();
temp.data = key;
temp.next = null ;
return temp;
}
static Node rearrangeEvenOdd(Node head)
{
if (head == null )
return null ;
Node odd = head;
Node even = head.next;
Node evenFirst = even;
while ( 1 == 1 )
{
if (odd == null || even == null ||
(even.next) == null )
{
odd.next = evenFirst;
break ;
}
odd.next = even.next;
odd = even.next;
if (odd.next == null )
{
even.next = null ;
odd.next = evenFirst;
break ;
}
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" ) ;
}
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);
}
}
|
Output:
Given Linked List
1->2->3->4->5->NULL
Modified Linked List
1->3->5->2->4->NULL
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Rearrange a linked list such that all even and odd positioned nodes are together for more details!
Last Updated :
22 Jun, 2022
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