Given a singly linked list, write a function to swap elements pairwise. For example, if the linked list is 1->2->3->4->5->6->7 then the function should change it to 2->1->4->3->6->5->7, and if the linked list is 1->2->3->4->5->6 then the function should change it to 2->1->4->3->6->5
This problem has been discussed here. The solution provided there swaps data of nodes. If data contains many fields, there will be many swap operations. So changing links is a better idea in general. Following is the implementation that changes links instead of swapping data.
// Java program to swap elements of // linked list by changing links class LinkedList
{ static Node head;
static class Node
{
int data;
Node next;
Node( int d)
{
data = d;
next = null ;
}
}
/* Function to pairwise swap elements
of a linked list */
Node pairWiseSwap(Node node)
{
// If linked list is empty or
// there is only one node in list
if (node == null ||
node.next == null )
{
return node;
}
// Initialize previous and
// current pointers
Node prev = node;
Node curr = node.next;
// Change head before proceeding
node = curr;
// Traverse the list
while ( true )
{
Node next = curr.next;
// Change next of current as
// previous node
curr.next = prev;
// If next NULL or next is the
// last node
if (next == null ||
next.next == null )
{
prev.next = next;
break ;
}
// Change next of previous to
// next next
prev.next = next.next;
// Update previous and curr
prev = next;
curr = prev.next;
}
return node;
}
/* Function to print nodes in a
given linked list */
void printList(Node node)
{
while (node != null )
{
System.out.print(node.data +
" " );
node = node.next;
}
}
// Driver code
public static void main(String[] args)
{
/* The constructed linked list is:
1->2->3->4->5->6->7 */
LinkedList list = new LinkedList();
list.head = new Node( 1 );
list.head.next = new Node( 2 );
list.head.next.next = new Node( 3 );
list.head.next.next.next =
new Node( 4 );
list.head.next.next.next.next =
new Node( 5 );
list.head.next.next.next.next.next =
new Node( 6 );
list.head.next.next.next.next.next.next =
new Node( 7 );
System.out.println(
"Linked list before calling pairwiseSwap() " );
list.printList(head);
Node st = list.pairWiseSwap(head);
System.out.println( "" );
System.out.println(
"Linked list after calling pairwiseSwap() " );
list.printList(st);
System.out.println( "" );
}
} // This code is contributed by Mayank Jaiswal |
Output:
Linked list before calling pairWiseSwap() 1 2 3 4 5 6 7 Linked list after calling pairWiseSwap() 2 1 4 3 6 5 7
Time Complexity: The time complexity of the above program is O(n) where n is the number of nodes in a given linked list. The while loop does a traversal of the given linked list.
Auxiliary Space: O(1)
Following is the recursive implementation of the same approach. We change the first two nodes and recur for the remaining list. Thanks to geek and omer salem for suggesting this method.
// Java program to swap elements of // linked list by changing links class LinkedList
{ static Node head;
static class Node
{
int data;
Node next;
Node( int d)
{
data = d;
next = null ;
}
}
/* Function to pairwise swap elements
of a linked list. It returns head
of the modified list, so return
value of this node must be assigned */
Node pairWiseSwap(Node node)
{
// Base Case: The list is empty or
// has only one node
if (node == null ||
node.next == null )
{
return node;
}
// Store head of list after two
// nodes
Node remaining = node.next.next;
// Change head
Node newhead = node.next;
// Change next of second node
node.next.next = node;
// Recur for remaining list and change
// next of head
node.next = pairWiseSwap(remaining);
// Return new head of modified list
return newhead;
}
/* Function to print nodes in a
given linked list */
void printList(Node node)
{
while (node != null )
{
System.out.print(node.data +
" " );
node = node.next;
}
}
// Driver code
public static void main(String[] args)
{
/* The constructed linked list is:
1->2->3->4->5->6->7 */
LinkedList list = new LinkedList();
list.head = new Node( 1 );
list.head.next = new Node( 2 );
list.head.next.next = new Node( 3 );
list.head.next.next.next =
new Node( 4 );
list.head.next.next.next.next =
new Node( 5 );
list.head.next.next.next.next.next =
new Node( 6 );
list.head.next.next.next.next.next.next =
new Node( 7 );
System.out.println(
"Linked list before calling pairwiseSwap() " );
list.printList(head);
head = list.pairWiseSwap(head);
System.out.println( "" );
System.out.println(
"Linked list after calling pairwiseSwap() " );
list.printList(head);
System.out.println( "" );
}
} |
Output:
Linked list before calling pairWiseSwap() 1 2 3 4 5 6 7 Linked list after calling pairWiseSwap() 2 1 4 3 6 5 7
Time Complexity: O(n)
Auxiliary Space: O(n)
Please refer complete article on Pairwise swap elements of a given linked list by changing links for more details!