Java Program For Swapping Nodes In A Linked List Without Swapping Data

Given a linked list and two keys in it, swap nodes for two given keys. Nodes should be swapped by changing links. Swapping data of nodes may be expensive in many situations when data contains many fields. 

It may be assumed that all keys in the linked list are distinct.

Examples:

Input : 10->15->12->13->20->14,  x = 12, y = 20
Output: 10->15->20->13->12->14

Input : 10->15->12->13->20->14,  x = 10, y = 20
Output: 20->15->12->13->10->14

Input : 10->15->12->13->20->14,  x = 12, y = 13
Output: 10->15->13->12->20->14

This may look a simple problem, but is an interesting question as it has the following cases to be handled. 

1. x and y may or may not be adjacent.
2. Either x or y may be a head node.
3. Either x or y may be the last node.
4. x and/or y may not be present in the linked list.

How to write a clean working code that handles all the above possibilities.

The idea is to first search x and y in the given linked list. If any of them is not present, then return. While searching for x and y, keep track of current and previous pointers. First change next of previous pointers, then change next of current pointers. 

Below is the implementation of the above approach. 

Java

// Java program to swap two given nodes
// of a linked list
class Node
{
    int data;
    Node next;
    Node(int d)
    {
        data = d;
        next = null;
    }
}

class LinkedList
{
    // head of list
    Node head;

    /* Function to swap Nodes x and y in
       linked list by changing links */
    public void swapNodes(int x, int y)
    {
        // Nothing to do if x and y
        // are same
        if (x == y)
            return;

        // Search for x (keep track of
        // prevX and CurrX)
        Node prevX = null, currX = head;
        while (currX != null &&
               currX.data != x)
        {
            prevX = currX;
            currX = currX.next;
        }

        // Search for y (keep track of
        // prevY and currY)
        Node prevY = null, currY = head;
        while (currY != null &&
               currY.data != y)
        {
            prevY = currY;
            currY = currY.next;
        }

        // If either x or y is not present,
        // nothing to do
        if (currX == null || currY == null)
            return;

        // If x is not head of linked list
        if (prevX != null)
            prevX.next = currY;
        else // make y the new head
            head = currY;

        // If y is not head of linked list
        if (prevY != null)
            prevY.next = currX;
        else // make x the new head
            head = currX;

        // Swap next pointers
        Node temp = currX.next;
        currX.next = currY.next;
        currY.next = temp;
    }

    // Function to add Node at
    // beginning of list.
    public void push(int new_data)
    {
        // 1. alloc the Node and put the data
        Node new_Node = new Node(new_data);

        // 2. Make next of new Node as head
        new_Node.next = head;

        // 3. Move the head to point to new Node
        head = new_Node;
    }

    /* This function prints contents of
       linked list starting from the given Node */
    public void printList()
    {
        Node tNode = head;
        while (tNode != null)
        {
            System.out.print(tNode.data + " ");
            tNode = tNode.next;
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        LinkedList llist = new LinkedList();

        /* The constructed linked list is:
           1->2->3->4->5->6->7 */
        llist.push(7);
        llist.push(6);
        llist.push(5);
        llist.push(4);
        llist.push(3);
        llist.push(2);
        llist.push(1);

        System.out.print(
               "Linked list before calling swapNodes() ");
        llist.printList();
        llist.swapNodes(4, 3);
        System.out.print(
               "Linked list after calling swapNodes() ");
        llist.printList();
    }
}
// This code is contributed by Rajat Mishra

Output:

Linked list before calling swapNodes() 1 2 3 4 5 6 7 
Linked list after calling swapNodes() 1 2 4 3 5 6 7 

Time Complexity: O(n)

Auxiliary Space: O(1)

Optimizations: The above code can be optimized to search x and y in single traversal. Two loops are used to keep program simple.

Simpler approach:

Java

// Java program to swap two given nodes
// of a linked list
public class Solution
{
    // Represent a node of the
    // singly linked list
    class Node
    {
        int data;
        Node next;

        public Node(int data)
        {
            this.data = data;
            this.next = null;
        }
    }

    // Represent the head and tail
    // of the singly linked list
    public Node head = null;
    public Node tail = null;

    // addNode() will add a new node
    // to the list
    public void addNode(int data)
    {
        // Create a new node
        Node newNode = new Node(data);

        // Checks if the list is empty
        if (head == null)
        {
            // If list is empty, both head and
            // tail will point to new node
            head = newNode;
            tail = newNode;
        }
        else
        {
            // newNode will be added after tail
            // such that tail's next will point
            // to newNode
            tail.next = newNode;
            // newNode will become new tail of
            // the list
            tail = newNode;
        }
    }

    // swap() will swap the given two nodes
    public void swap(int n1, int n2)
    {
        Node prevNode1 = null, prevNode2 = null,
             node1 = head, node2 = head;

        // Checks if list is empty
        if (head == null)
        {
            return;
        }

        // If n1 and n2 are equal, then
        // list will remain the same
        if (n1 == n2)
            return;

        // Search for node1
        while (node1 != null &&
               node1.data != n1)
        {
            prevNode1 = node1;
            node1 = node1.next;
        }

        // Search for node2
        while (node2 != null &&
               node2.data != n2)
        {
            prevNode2 = node2;
            node2 = node2.next;
        }

        if (node1 != null &&
            node2 != null)
        {
            // If previous node to node1 is not
            // null then, it will point to node2
            if (prevNode1 != null)
                prevNode1.next = node2;
            else
                head = node2;

            // If previous node to node2 is not
            // null then, it will point to node1
            if (prevNode2 != null)
                prevNode2.next = node1;
            else
                head = node1;

            // Swaps the next nodes of node1
            // and node2
            Node temp = node1.next;
            node1.next = node2.next;
            node2.next = temp;
        }
        else
        {
            System.out.println("Swapping is not possible");
        }
    }

    // display() will display all the
    // nodes present in the list
    public void display()
    {
        // Node current will point to head
        Node current = head;

        if (head == null)
        {
            System.out.println("List is empty");
            return;
        }
        while (current != null)
        {
            // Prints each node by incrementing
            // pointer
            System.out.print(current.data + " ");
            current = current.next;
        }
        System.out.println();
    }

    public static void main(String[] args)
    {
        Solution sList = new Solution();

        // Add nodes to the list
        sList.addNode(1);
        sList.addNode(2);
        sList.addNode(3);
        sList.addNode(4);
        sList.addNode(5);
        sList.addNode(6);
        sList.addNode(7);

        System.out.println("Original list: ");
        sList.display();

        // Swaps the node 2 with node 5
        sList.swap(6, 1);

        System.out.println(
               "List after swapping nodes: ");
        sList.display();
    }
}

Output:

Linked list before calling swapNodes() 1 2 3 4 5 6 7 
Linked list after calling swapNodes() 6 2 3 4 5 1 7 

Time Complexity: O(n)

Auxiliary Space: O(1)

Please refer complete article on Swap nodes in a linked list without swapping data for more details!