Given a singly linked list, find middle of the linked list and set middle node of the linked list at beginning of the linked list.
Examples:
Input: 1 2 3 4 5
Output: 3 1 2 4 5
Input: 1 2 3 4 5 6
Output: 4 1 2 3 5 6

The idea is to first find middle of a linked list using two pointers, first one moves one at a time and second one moves two at a time. When second pointer reaches end, first reaches middle. We also keep track of previous of first pointer so that we can remove middle node from its current position and can make it head.
Java
public class GFG
{
static class Node
{
int data;
Node next;
Node( int data)
{
this .data = data;
next = null ;
}
}
static Node head;
static void setMiddleHead()
{
if (head == null )
return ;
Node one_node = head;
Node two_node = head;
Node prev = null ;
while (two_node != null &&
two_node.next != null )
{
prev = one_node;
two_node = two_node.next.next;
one_node = one_node.next;
}
prev.next = prev.next.next;
one_node.next = head;
head = one_node;
}
static void push( int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
static void printList(Node ptr)
{
while (ptr != null )
{
System.out.print(ptr.data + " " );
ptr = ptr.next;
}
System.out.println();
}
public static void main(String args[])
{
head = null ;
int i;
for (i = 5 ; i > 0 ; i--)
push(i);
System.out.print( " list before: " );
printList(head);
setMiddleHead();
System.out.print( " list After: " );
printList(head);
}
}
|
Output:
list before: 1 2 3 4 5
list After : 3 1 2 4 5
Time Complexity: O(n) where n is the total number of nodes in the linked list.
Space Complexity: O(1) since using constant space
Please refer complete article on Make middle node head in a linked list for more details!
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Last Updated :
02 Feb, 2023
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