Swap Kth node from beginning with Kth node from end in a Doubly Linked List
Prerequisites: Doubly Linked List
Given a doubly-linked list, the task is to swap Kth node from the beginning with Kth node from the ending.
Note: Please note here the nodes are swapped and not the data in the nodes.
Examples:
Input: DLL = 1 <-> 2 <-> 3 <-> 4 <-> 5 <-> 6, K = 3
Output: 1 2 4 3 5 6
Explanation:
Third node from the beginning(3) is swapped with third node from the ending(4).Input: DLL = 1 <-> 2 <-> 3 <-> 4 <-> 5, K = 1
Output: 5 2 3 4 1
Approach: The idea is to traverse to the Kth element from the beginning and Kth node from the ending and change the previous and next pointers. Let K1 be the Kth node from beginning and K2 be Kth node from ending. Then:
- The previous node to K2 has to be changed to the previous node of K1.
- The next node to K2 has to be changed to the next node of K1.
- The previous node to K1 has to be changed to the previous node of K2.
- The next node to K1 has to be changed to the next node of K2.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Node structure of the doubly linked list struct Node { int data; Node* next; Node* previous; }; // Class of the doubly linked list class GFG { // head of the doubly linked list Node* head; // tail of the doubly linked list Node* tail; public : // Default constructor GFG() { head = tail = NULL; } // Function to return the head of the // doubly linked list Node* getHead() { return head; } // Function to set the head of the // doubly linked list void setHead(Node* head) { this ->head = head; } // Function to return the tail of the // doubly linked list Node* getTail() { return tail; } // Function to set the tail of the // doubly linked list void setTail(Node* tail) { this ->tail = tail; } // Function to replace Kth node from // beginning with Kth node from end void swapNode(Node* headReference, Node* tailReference, int k) { // If K is 1, then the first node // has to be swapped with the // last node in the doubly linked list if (k == 1) { swapFirstAndLast(headReference, tailReference); return ; } // If k is N, then the last node // has to be swapped with the // first node in the doubly linked list int nodeCount = getCount(headReference); if (k == nodeCount) { swapFirstAndLast(headReference, tailReference); return ; } // If the K<sup>th</sup> node from // the beginning and K<sup>th</sup> node // from the ending are same if (2 * k - 1 == nodeCount) { return ; } // fNode represents K<sup>th</sup> node // from the beginning Node* fNode = headReference; for ( int i = 1; i < k; i++) { fNode = fNode->next; } Node* fNodePrevious = fNode->previous; Node* fNodeNext = fNode->next; // sNode represents K<sup>th</sup> node // from the ending Node* sNode = tailReference; for ( int i = 1; i < k; i++) { sNode = sNode->previous; } Node* sNodePrevious = sNode->previous; Node* sNodeNext = sNode->next; // Checking if any of the pointers is null // and interchanging the pointers if (fNodePrevious != NULL && sNode != NULL) { fNodePrevious->next = sNode; sNode->previous = fNodePrevious; sNode->next = fNodeNext; fNodeNext->previous = sNode; } if (sNodePrevious != NULL && sNodeNext != NULL) { sNodeNext->previous = fNode; fNode->next = sNodeNext; sNodePrevious ->next = fNode; fNode->previous = sNodePrevious; } } // Function to swap the first and // last node in the doubly linked list void swapFirstAndLast(Node* headReference, Node* tailReference) { Node* headRef = headReference; Node* tailRef = tailReference; headReference = headReference->next; tailReference = tailReference->previous; tailReference->next = headRef; headRef->previous = tailReference; headRef->next = NULL; this ->setTail(tailReference ->next); headReference->previous = tailRef; tailRef->next = headReference; tailRef->previous = NULL; this ->setHead(headReference ->previous); } // Function to return the number of nodes // in the linked list int getCount(Node* headReference) { int nodeCount = 0; while (headReference != NULL) { nodeCount++; headReference = headReference->next; } return nodeCount; } // Function to print the Linked List void printList(Node* headReference) { if (headReference == NULL) { cout << "Doubly linked list is empty" ; return ; } else { while (headReference != NULL) { cout << headReference->data << " " ; headReference = headReference->next; } } } // Function to insert a node at // the end of the doubly linked list void push( int data) { Node* newNode = new Node; newNode->data = data; newNode->next = NULL; newNode->previous = NULL; if (head == NULL) { head = tail = newNode; } else { tail->next = newNode; newNode->previous = tail; tail = newNode; } } }; // Driver code int main() { // Creating an object for the class GFG list; // Adding data to the linked list list.push(1); list.push(2); list.push(3); list.push(4); list.push(5); // Calling the function int K = 2; list.swapNode(list.getHead(), list.getTail(), K); list.printList(list.getHead()); return 0; } |
Java
// Java implementation of the approach public class GFG { // Doubly Linked List implementation private class Node { private int data; private Node next; private Node previous; public Node( int data, Node next, Node previous) { this .data = data; this .next = next; this .previous = previous; } public int getData() { return data; } public void setData( int data) { this .data = data; } public Node getNext() { return next; } public void setNext(Node next) { this .next = next; } public Node getPrevious() { return previous; } public void setPrevious(Node previous) { this .previous = previous; } } private Node head; private Node tail; public GFG() { this .head = null ; this .tail = null ; } public Node getHead() { return head; } public void setHead(Node head) { this .head = head; } public Node getTail() { return tail; } public void setTail(Node tail) { this .tail = tail; } // Function to replace Kth node from // beginning with Kth node from end public void swapNode(Node headReference, Node tailReference, int k) { // If K is 1, then the first node // has to be swapped with the // last node in the doubly linked list if (k == 1 ) { swapFirstAndLast(headReference, tailReference); return ; } // If k is N, then the last node // has to be swapped with the // first node in the doubly linked list int nodeCount = getCount(headReference); if (k == nodeCount) { swapFirstAndLast(headReference, tailReference); return ; } // If the K<sup>th</sup> node from // the beginning and K<sup>th</sup> node // from the ending are same if ( 2 * k - 1 == nodeCount) { return ; } // fNode represents K<sup>th</sup> node // from the beginning Node fNode = headReference; for ( int i = 1 ; i < k; i++) { fNode = fNode.getNext(); } Node fNodePrevious = fNode.getPrevious(); Node fNodeNext = fNode.getNext(); // sNode represents K<sup>th</sup> node // from the ending Node sNode = tailReference; for ( int i = 1 ; i < k; i++) { sNode = sNode.getPrevious(); } Node sNodePrevious = sNode.getPrevious(); Node sNodeNext = sNode.getNext(); // Checking if any of the pointers is null // and interchanging the pointers if (fNodePrevious != null && sNode != null ) { fNodePrevious.setNext(sNode); sNode.setPrevious(fNodePrevious); sNode.setNext(fNodeNext); fNodeNext.setPrevious(sNode); } if (sNodePrevious != null && sNodeNext != null ) { sNodeNext.setPrevious(fNode); fNode.setNext(sNodeNext); sNodePrevious.setNext(fNode); fNode.setPrevious(sNodePrevious); } } // Function to swap the first and // last node in the doubly linked list private void swapFirstAndLast( Node headReference, Node tailReference) { Node headRef = headReference; Node tailRef = tailReference; headReference = headReference.getNext(); tailReference = tailReference.getPrevious(); tailReference.setNext(headRef); headRef.setPrevious(tailReference); headRef.setNext( null ); this .setTail(tailReference.getNext()); headReference.setPrevious(tailRef); tailRef.setNext(headReference); tailRef.setPrevious( null ); this .setHead(headReference .getPrevious()); } // Function to return the number of nodes // in the linked list private int getCount(Node headReference) { int nodeCount = 0 ; while (headReference != null ) { nodeCount++; headReference = headReference .getNext(); } return nodeCount; } // Function to print the Linked List public void printList(Node headReference) { if (headReference == null ) { System.out.println( "Doubly linked list is empty" ); return ; } else { while (headReference != null ) { System.out.print( headReference.getData() + " " ); headReference = headReference.getNext(); } } } // Function to insert a node at // the end of the doubly linked list public void push( int data) { Node newNode = new Node(data, null , null ); if (head == null ) { head = tail = newNode; } else { tail.setNext(newNode); newNode.setPrevious(tail); tail = newNode; } } // Driver code public static void main(String[] args) { // Creating an object for the class GFG list = new GFG(); // Adding data to the linked list list.push( 1 ); list.push( 2 ); list.push( 3 ); list.push( 4 ); list.push( 5 ); // Calling the function int K = 2 ; list.swapNode(list.getHead(), list.getTail(), K); list.printList(list.getHead()); } } |
Python3
# Python implementation of the approach # Node structure of the doubly linked list class Node: def __init__( self , data): self .data = data self . next = None self .previous = None def getData( self ): return self .data def setData( self , data): self .data = data def getNext( self , ): return next def setNext( self , next ): self . next = next def getPrevious( self ): return previous def setPrevious( self , previous): self .previous = previous # Class of the doubly linked list class GFG: # head of the doubly linked list def __init__( self ): self .head = None # tail of the doubly linked list self .tail = None # Function to return the head of the doubly linked list def getHead( self ): return self .head # Function to set the head of the doubly linked list def setHead( self , head): self .head = head # Function to return the tail of the doubly linked list def getTail( self ): return self .tail # Function to set the tail of the doubly linked list def setTail( self , tail): self .tail = tail # Function to replace Kth node from beginning with Kth node from end def swapNode( self , headReference, tailReference, k): # If K is 1, then the first node has to be swapped with the last node in the doubly linked list if k = = 1 : self .swapFirstAndLast(headReference, tailReference) return # If k is N, then the last node has to be swapped with the first node in the doubly linked list nodeCount = self .getCount(headReference) if k = = nodeCount: self .swapFirstAndLast(headReference, tailReference) return # If the Kth node from the beginning and Kth node from the ending are same if 2 * k - 1 = = nodeCount: return # fNode represents Kth node from the beginning fNode = headReference for i in range ( 1 , k): fNode = fNode. next fNodePrevious = fNode.previous fNodeNext = fNode. next # sNode represents Kth node from the ending sNode = tailReference for i in range ( 1 , k): sNode = sNode.previous sNodePrevious = sNode.previous sNodeNext = sNode. next # Checking if any of the pointers is null and interchanging the pointers if fNodePrevious ! = None and sNode ! = None : fNodePrevious. next = sNode sNode.previous = fNodePrevious sNode. next = fNodeNext fNodeNext.previous = sNode if sNodePrevious ! = None and sNodeNext ! = None : sNodeNext.previous = fNode fNode. next = sNodeNext sNodePrevious. next = fNode fNode.previous = sNodePrevious # Function to swap the first and last node in the doubly linked list def swapFirstAndLast( self , headReference, tailReference): headRef = headReference tailRef = tailReference headReference = headReference. next tailReference = tailReference.previous tailReference. next = headRef headRef.previous = tailReference headRef. next = None self .setTail(tailReference. next ) headReference.previous = tailRef tailRef. next = headReference tailRef.previous = None self .setHead(headReference.previous) # Function to return the number of nodes in the linked list def getCount( self , headReference): nodeCount = 0 while headReference ! = None : nodeCount + = 1 headReference = headReference. next return nodeCount # Function to insert a node at # the end of the doubly linked list def push( self , data): newNode = Node(data) if ( self .head = = None ): self .head = self .tail = newNode else : self .tail.setNext(newNode) newNode.setPrevious( self .tail) self .tail = newNode def printList( self , headReference): if headReference = = None : print ( "Doubly linked list is empty" ) return while headReference: print (headReference.data,end = ' ' ) headReference = headReference. next print () # Driver code # Creating an object for the class list = GFG() # Adding data to the linked list list .push( 1 ) list .push( 2 ) list .push( 3 ) list .push( 4 ) list .push( 5 ) # Calling the function K = 2 list .swapNode( list .getHead(), list .getTail(), K) list .printList( list .getHead()) |
C#
// C# implementation of the approach using System; public class GFG { // Doubly Linked List implementation private class Node { private int data; private Node next; private Node previous; public Node( int data, Node next, Node previous) { this .data = data; this .next = next; this .previous = previous; } public int getData() { return data; } public void setData( int data) { this .data = data; } public Node getNext() { return next; } public void setNext(Node next) { this .next = next; } public Node getPrevious() { return previous; } public void setPrevious(Node previous) { this .previous = previous; } } private Node head; private Node tail; public GFG() { this .head = null ; this .tail = null ; } Node getHead() { return head; } void setHead(Node head) { this .head = head; } Node getTail() { return tail; } void setTail(Node tail) { this .tail = tail; } // Function to replace Kth node from // beginning with Kth node from end void swapNode(Node headReference, Node tailReference, int k) { // If K is 1, then the first node // has to be swapped with the // last node in the doubly linked list if (k == 1) { swapFirstAndLast(headReference, tailReference); return ; } // If k is N, then the last node // has to be swapped with the // first node in the doubly linked list int nodeCount = getCount(headReference); if (k == nodeCount) { swapFirstAndLast(headReference, tailReference); return ; } // If the K<sup>th</sup> node from // the beginning and K<sup>th</sup> node // from the ending are same if (2 * k - 1 == nodeCount) { return ; } // fNode represents K<sup>th</sup> node // from the beginning Node fNode = headReference; for ( int i = 1; i < k; i++) { fNode = fNode.getNext(); } Node fNodePrevious = fNode.getPrevious(); Node fNodeNext = fNode.getNext(); // sNode represents K<sup>th</sup> node // from the ending Node sNode = tailReference; for ( int i = 1; i < k; i++) { sNode = sNode.getPrevious(); } Node sNodePrevious = sNode.getPrevious(); Node sNodeNext = sNode.getNext(); // Checking if any of the pointers is null // and interchanging the pointers if (fNodePrevious != null && sNode != null ) { fNodePrevious.setNext(sNode); sNode.setPrevious(fNodePrevious); sNode.setNext(fNodeNext); fNodeNext.setPrevious(sNode); } if (sNodePrevious != null && sNodeNext != null ) { sNodeNext.setPrevious(fNode); fNode.setNext(sNodeNext); sNodePrevious.setNext(fNode); fNode.setPrevious(sNodePrevious); } } // Function to swap the first and // last node in the doubly linked list private void swapFirstAndLast( Node headReference, Node tailReference) { Node headRef = headReference; Node tailRef = tailReference; headReference = headReference.getNext(); tailReference = tailReference.getPrevious(); tailReference.setNext(headRef); headRef.setPrevious(tailReference); headRef.setNext( null ); this .setTail(tailReference.getNext()); headReference.setPrevious(tailRef); tailRef.setNext(headReference); tailRef.setPrevious( null ); this .setHead(headReference .getPrevious()); } // Function to return the number of nodes // in the linked list private int getCount(Node headReference) { int nodeCount = 0; while (headReference != null ) { nodeCount++; headReference = headReference .getNext(); } return nodeCount; } // Function to print the Linked List void printList(Node headReference) { if (headReference == null ) { Console.WriteLine( "Doubly linked list is empty" ); return ; } else { while (headReference != null ) { Console.Write( headReference.getData() + " " ); headReference = headReference.getNext(); } } } // Function to insert a node at // the end of the doubly linked list void Push( int data) { Node newNode = new Node(data, null , null ); if (head == null ) { head = tail = newNode; } else { tail.setNext(newNode); newNode.setPrevious(tail); tail = newNode; } } // Driver code public static void Main(String[] args) { // Creating an object for the class GFG list = new GFG(); // Adding data to the linked list list.Push(1); list.Push(2); list.Push(3); list.Push(4); list.Push(5); // Calling the function int K = 2; list.swapNode(list.getHead(), list.getTail(), K); list.printList(list.getHead()); } } // This code is contributed by 29AjayKumar |
1 4 3 2 5
Time Complexity: O(K), As we are traversing till Kth element from tail and head of the list and then changing the links which is a O(1) operation.
Auxiliary Space: O(1), As constant extra space is used.
Method 2: Without swapping the elements and without using temporary node.
Approach: There are 3 cases in order to swap the nodes.
- Swapping the first and the last nodes (k = 1)
- Swapping the ordinary Kth node from the beginning and Kth node from the end.
- Swapping middle nodes
Case 1: Swap first and last nodes (k = 1)
Steps:
- Make the list as a circular linked list
- Change the previous pointer of the first node to the last but one node (20 in example figure)
- Change the next pointer of last but one node to the last node. In this case it will be 60.
- After swapping, make the head as the first node.
Consider p and q are the nodes which are to be swapped, head = q; //change head pointer to point to head node last = p; //change last pointer to point to last node

swapping first and last nodes
Case 2: Swapping the ordinary Kth node from the beginning and Kth node from the end.
Steps:
- Let us consider K = 2. So the nodes to be swapped or interchanged are 20 and 50 as show in the figure.
- Make both the first and next pointers of the nodes which are to be swapped to point to the previous nodes. To do this, we need to change the links of the previous nodes to point to the node which is after the node to be swapped.
Consider the nodes to be swapped are p and q: //Change the link of the next pointer of the previous node to point to //the next node of to be swapped node. q.first.next = q.next; p.first.next = p.next; // Same procedure for the other node //Make sure to change the previous/first pointer of the next node to //point to the previous of to be swapped node. q.next.first = q.first; p.next.first = p.first; //Both the first and next pointers points to the previous node as shown in the below figure. q.next = q.first; p.next = p.first;
3. Interchange the pointers of one node to be swapped nodes with the other to be swapped node. (step 3 denotes the figure after interchanging).
4. Make the required changes in the links in order to make it as a complete list.

General case
Case 3: Swapping the middle nodes
Steps:
- This case is same as the case 2 and the only change is, the nodes which are to be swapped are middle nodes. So both of them are together (side-by-side).
- Consider p is the first node to be swapped and q is the second node to be swapped.
- Point the next pointer of previous node of p to the next node of q. This step is done to omit p and q nodes.
- In the same way, point the first pointer of next node of q to the previous node of q.
- Change the links of p and q so that both the nodes points to the previous node of p (step2 in the below figure).
- Make the links of p and q accordingly to make the nodes swap their positions.

To swap middle nodes
Implementation:
C++
// CPP program to swap Kth node from beginning with // the Kth node from the end without using temporary // node and without swapping the data #include <bits/stdc++.h> using namespace std; // class Node class Node { public : int data; Node *first, *next; Node( int data) { this ->data = data; first = NULL; next = NULL; } }; // function for inserting new node at the // end of the list using last pointer void AddLast( int data, Node*& head, Node*& last) { Node* temp = new Node(data); if (!head) { head = temp; last = temp; } else { last->next = temp; temp->first = last; last = temp; } } // function for printing the doubly linked list void printList(Node* head) { Node* p = head; while (p) { cout << p->data << "<->" ; p = p->next; } cout << "NULL" << endl; } // function for swapping Kth node from // beginning with Kth node from the end void swapKthNodes(Node*& head, Node*& last, int k) { int count = 1; Node *p = head, *q = last; // case 1: to swap the start and end nodes // case 1 figure if (k == 1) { q->first->next = p; p->first = q->first; q->next = p->next; p->next->first = q; // change these links to NULL to the break circular // link p->next = NULL; q->first = NULL; head = q; last = p; } else { while (p != NULL && q != NULL && count < k) { count++; p = p->next; q = q->first; } // case 3: if the nodes to be swapped are the middle // nodes given in the figure if (p->next == q) { p->first->next = p->next->next; q->next->first = p->first; p->next = p->first; q->first = q->next = p->first; q->next = p; p->next->next->first = p; p->next = p->first->next; p->first->next = q; p->first = q; } // case 2: other than middle nodes // given in case 2 figure else { q->first->next = q->next; q->next->first = q->first; q->next = q->first; p->first->next = p->next; p->next->first = p->first; p->next = p->first; p->first = q->first; q->first = p->next; p->next = p->first; q->next = q->first; q->next = q->next->next; q->first->next = q; q->next->first = q; p->next = p->next->next; p->first->next = p; p->next->first = p; } } } // Driver function int main() { // head pointer for pointing to start of the linked list // last pointer for pointing to last node of the linked // list Node *head = NULL, *last = NULL; // function calls for inserting // at the end of the list AddLast(10, head, last); AddLast(20, head, last); AddLast(30, head, last); AddLast(40, head, last); AddLast(50, head, last); AddLast(60, head, last); cout << "Before swapping:" << endl; // print list before swapping the nodes printList(head); cout << endl; // function call for swapping Kth nodes swapKthNodes(head, last, 1); cout << "After swapping nodes for k = 1:" << endl; // print list after swapping the nodes printList(head); cout << endl; swapKthNodes(head, last, 2); cout << "After swapping nodes for k = 2:" << endl; printList(head); cout << endl; swapKthNodes(head, last, 3); cout << "After swapping nodes for k = 3 (middle):" << endl; printList(head); cout << endl; return 0; } // This code is contributed Tapesh(tapeshdua420) |
Java
// Java program to swap Kth node from beginning with // the Kth node from the end without using temporary // node and without swapping the data public class GFG { // head pointer for pointing to start of the linked list // last pointer for pointing to last node of the linked list Node head = null ,last = null ; // class Node class Node{ int data; Node first,next; Node( int data){ this .data = data; first = null ; next = null ; } } // function for inserting new node at the // end of the list using last pointer void AddLast( int data) { Node temp = new Node(data); if (head == null ) { head = temp; last = temp; } else { last.next = temp; temp.first = last; last = temp; } } // function for printing the doubly linked list void printList() { Node p = head; while (p!= null ) { System.out.print(p.data+ "<->" ); p = p.next; } System.out.print( "null" ); System.out.println(); } // function for swapping Kth node from // beginning with Kth node from the end void swapKthNodes( int k) { int count = 1 ; Node p = head, q = last; // case 1: to swap the start and end nodes // case 1 figure if (k == 1 ) { q.first.next = p; p.first = q.first; q.next = p.next; p.next.first = q; // change these links to null to the break circular link p.next = null ; q.first = null ; head = q; last = p; } else { while (p != null && q != null && count < k) { count++; p = p.next; q = q.first; } // case 3: if the nodes to be swapped are the middle nodes // given in the figure if (p.next == q) { p.first.next = p.next.next; q.next.first = p.first; p.next = p.first; q.first = q.next = p.first; q.next = p; p.next.next.first = p; p.next = p.first.next; p.first.next = q; p.first = q; } // case 2: other than middle nodes // given in case 2 figure else { q.first.next = q.next; q.next.first = q.first; q.next = q.first; p.first.next = p.next; p.next.first = p.first; p.next = p.first; p.first = q.first; q.first = p.next; p.next = p.first; q.next = q.first; q.next = q.next.next; q.first.next = q; q.next.first = q; p.next = p.next.next; p.first.next = p; p.next.first = p; } } } // Driver function public static void main(String args[]) { // class object GFG list = new GFG(); // function calls for inserting // at the end of the list list.AddLast( 10 ); list.AddLast( 20 ); list.AddLast( 30 ); list.AddLast( 40 ); list.AddLast( 50 ); list.AddLast( 60 ); System.out.println( "Before swapping:" ); // print list before swapping the nodes list.printList(); System.out.println(); // function call for swapping Kth nodes list.swapKthNodes( 1 ); System.out.println( "After swapping nodes for k = 1:" ); // print list after swapping the nodes list.printList(); System.out.println(); list.swapKthNodes( 2 ); System.out.println( "After swapping nodes for k = 2:" ); list.printList(); System.out.println(); list.swapKthNodes( 3 ); System.out.println( "After swapping nodes for k = 3 (middle):" ); list.printList(); System.out.println(); } } // This code is contributed by Likhita AVL |
Python3
# Python program to swap Kth node from beginning with # the Kth node from the end without using temporary # node and without swapping the data # class Node class Node: def __init__( self , data): self .data = data self .first = None self . next = None class GFG: # head pointer for pointing to start of the linked list # last pointer for pointing to last node of the linked list def __init__( self ): self .head = None self .last = None # function for inserting new node at the # end of the list using last pointer def AddLast( self , data): temp = Node(data) if self .head = = None : self .head = temp self .last = temp else : self .last. next = temp temp.first = self .last self .last = temp # function for printing the doubly linked list def printList( self ): p = self .head while p ! = None : print (p.data, "<->" , end = "") p = p. next print ( "null" ) # function for swapping Kth node from # beginning with Kth node from the end def swapKthNodes( self , k): count = 1 p = self .head q = self .last # case 1: to swap the start and end nodes # case 1 figure if k = = 1 : q.first. next = p p.first = q.first q. next = p. next p. next .first = q # change these links to null to the break circular link p. next = None q.first = None self .head = q self .last = p else : while p ! = None and q ! = None and count < k: count + = 1 p = p. next q = q.first # case 3: if the nodes to be swapped are the middle nodes # given in the figure if p. next = = q: p.first. next = p. next . next q. next .first = p.first p. next = p.first q.first = q. next = p.first q. next = p p. next . next .first = p p. next = p.first. next p.first. next = q p.first = q # case 2: other than middle nodes # given in case 2 figure else : q.first. next = q. next q. next .first = q.first q. next = q.first p.first. next = p. next p. next .first = p.first p. next = p.first p.first = q.first q.first = p. next p. next = p.first q. next = q.first q. next = q. next . next q.first. next = q q. next .first = q p. next = p. next . next p.first. next = p p. next .first = p # Driver function if __name__ = = '__main__' : # class object list = GFG() # function calls for inserting # at the end of the list list .AddLast( 10 ) list .AddLast( 20 ) list .AddLast( 30 ) list .AddLast( 40 ) list .AddLast( 50 ) list .AddLast( 60 ) print ( "Before swapping:" ) # print list before swapping the nodes list .printList() print () # function call for swapping Kth nodes list .swapKthNodes( 1 ) print ( "After swapping nodes for k = 1:" ) # print list after swapping the nodes list .printList() print () list .swapKthNodes( 2 ) print ( "After swapping nodes for k = 2:" ) list .printList() print () list .swapKthNodes( 3 ) print ( "After swapping nodes for k = 3 (middle):" ) list .printList() print () # This code is contributed by Tapesh(tapeshuda420) |
C#
// C# program to swap Kth node from beginning with the Kth // node from the end without using temporary node and without // swapping the data using System; public class GFG { // class Node class Node { public int data; public Node first, next; public Node( int data) { this .data = data; first = null ; next = null ; } } // head pointer for pointing to start of the linked list // last pointer for pointing to last node of the linked // list Node head = null , last = null ; // function for inserting new node at the end of the // list using last pointer void AddLast( int data) { Node temp = new Node(data); if (head == null ) { head = temp; last = temp; } else { last.next = temp; temp.first = last; last = temp; } } // function for printing the doubly linked list void printList() { Node p = head; while (p != null ) { Console.Write(p.data + "<->" ); p = p.next; } Console.Write( "null" ); Console.WriteLine(); } // function for swapping Kth node from beginning with // Kth node from the end void swapKthNodes( int k) { int count = 1; Node p = head, q = last; // case 1: to swap the start and end nodes case 1 // figure if (k == 1) { q.first.next = p; p.first = q.first; q.next = p.next; p.next.first = q; // change these links to null to the break // circular link p.next = null ; q.first = null ; head = q; last = p; } else { while (p != null && q != null && count < k) { count++; p = p.next; q = q.first; } // case 3: if the nodes to be swapped are the // middle nodes given in the figure if (p.next == q) { p.first.next = p.next.next; q.next.first = p.first; p.next = p.first; q.first = q.next = p.first; q.next = p; p.next.next.first = p; p.next = p.first.next; p.first.next = q; p.first = q; } // case 2: other than middle nodes given in case // 2 figure else { q.first.next = q.next; q.next.first = q.first; q.next = q.first; p.first.next = p.next; p.next.first = p.first; p.next = p.first; p.first = q.first; q.first = p.next; p.next = p.first; q.next = q.first; q.next = q.next.next; q.first.next = q; q.next.first = q; p.next = p.next.next; p.first.next = p; p.next.first = p; } } } static public void Main() { GFG list = new GFG(); // function calls for inserting // at the end of the list list.AddLast(10); list.AddLast(20); list.AddLast(30); list.AddLast(40); list.AddLast(50); list.AddLast(60); Console.WriteLine( "Before swapping:" ); // print list before swapping the nodes list.printList(); Console.WriteLine(); // function call for swapping Kth nodes list.swapKthNodes(1); Console.WriteLine( "After swapping nodes for k = 1:" ); // print list after swapping the nodes list.printList(); Console.WriteLine(); list.swapKthNodes(2); Console.WriteLine( "After swapping nodes for k = 2:" ); list.printList(); Console.WriteLine(); list.swapKthNodes(3); Console.WriteLine( "After swapping nodes for k = 3 (middle):" ); list.printList(); Console.WriteLine(); } } // This code is contributed by lokesh (lokeshmvs21). |
Javascript
// Javascript code class Node { constructor(data) { this .data = data; this .first = null ; this .next = null ; } } // function for inserting new node at the // end of the list using last pointer function AddLast(data, head, last) { let temp = new Node(data); if (!head) { head = temp; last = temp; } else { last.next = temp; temp.first = last; last = temp; } return [head, last]; } // function for printing the doubly linked list function printList(head) { let p = head; while (p) { console.log(`${p.data}<->`); p = p.next; } console.log( "NULL" ); } // function for swapping Kth node from // beginning with Kth node from the end function swapKthNodes(head, last, k) { let count = 1; let p = head, q = last; // case 1: to swap the start and end nodes // case 1 figure if (k == 1) { q.first.next = p; p.first = q.first; q.next = p.next; p.next.first = q; // change these links to NULL to the break circular // link p.next = null ; q.first = null ; head = q; last = p; return [head, last]; } else { while (p != null && q != null && count < k) { count++; p = p.next; q = q.first; } // case 3: if the nodes to be swapped are the middle // nodes given in the figure if (p.next == q) { p.first.next = p.next.next; q.next.first = p.first; p.next = p.first; q.first = q.next = p.first; q.next = p; p.next.next.first = p; p.next = p.first.next; p.first.next = q; p.first = q; } // case 2: other than middle nodes // given in case 2 figure else { q.first.next = q.next; q.next.first = q.first; q.next = q.first; p.first.next = p.next; p.next.first = p.first; p.next = p.first; p.first = q.first; q.first = p.next; p.next = p.first; q.next = q.first; q.next = q.next.next; q.first.next = q; q.next.first = q; p.next = p.next.next; p.first.next = p; p.next.first = p; } } return [head, last]; } // Driver function // head pointer for pointing to start of the linked list // last pointer for pointing to last node of the linked // list let head = null , last = null ; // function calls for inserting // at the end of the list [head, last] = AddLast(10, head, last); [head, last] = AddLast(20, head, last); [head, last] = AddLast(30, head, last); [head, last] = AddLast(40, head, last); [head, last] = AddLast(50, head, last); [head, last] = AddLast(60, head, last); console.log( "Before swapping:" ); // print list before swapping the nodes printList(head); console.log(); // function call for swapping Kth nodes [head, last] = swapKthNodes(head, last, 1); console.log( "After swapping nodes for k = 1:" ); // print list after swapping the nodes printList(head); console.log(); [head, last] = swapKthNodes(head, last, 2); console.log( "After swapping nodes for k = 2:" ); printList(head); console.log(); [head, last] = swapKthNodes(head, last, 3); console.log( "After swapping nodes for k = 3 (middle):" ); printList(head); console.log(); // This code is contributed by akashish__ |
Before swapping: 10<->20<->30<->40<->50<->60<->NULL After swapping nodes for k = 1: 60<->20<->30<->40<->50<->10<->NULL After swapping nodes for k = 2: 60<->50<->30<->40<->20<->10<->NULL After swapping nodes for k = 3 (middle): 60<->50<->40<->30<->20<->10<->NULL
Time complexity: O(N) where N is number of nodes in linked list
Auxiliary Space: O(1)
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