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

• Last Updated : 30 Mar, 2022

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!

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