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

Last Updated : 09 Sep, 2024

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.

x and y may or may not be adjacent.
Either x or y may be a head node.
Either x or y may be the last node.
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.

JavaScript

// JavaScript program to swap two  // given nodes of a linked list class Node {     constructor(val) {         this.data = val;         this.next = null;     } }  // Head of list let head;  /* Function to swap Nodes x and y in     linked list by changing links */ function swapNodes(x, y) {     // Nothing to do if x and y      // are same     if (x == y)         return;      // Search for x (keep track of      // prevX and CurrX)     let prevX = null, currX = head;     while (currX != null &&         currX.data != x) {         prevX = currX;         currX = currX.next;     }      // Search for y (keep track of      // prevY and currY)     let 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     let temp = currX.next;     currX.next = currY.next;     currY.next = temp; }  // Function to add Node at beginning  // of list function push(new_data) {     // 1. alloc the Node and put the data     let 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 function printList() {     let tNode = head;     while (tNode != null) {         console.log(tNode.data);         tNode = tNode.next;     } }  // Driver code /* The constructed linked list is:    1->2->3->4->5->6->7 */ push(7); push(6); push(5); push(4); push(3); push(2); push(1);  console.log("Linked list before calling swapNodes()"); printList(); swapNodes(4, 3); console.log("Linked list after calling swapNodes()"); printList(); // This code is contributed by todaysgaurav

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

Complexity Analysis:

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:

JavaScript

// Javascript program to swap two given  // nodes of a linked list // Represent a node of the singly  // linked list class Node {     constructor(val) {         this.data = val;         this.next = null;     } }  // Represent the head and tail of  // the singly linked list let head = null; let tail = null;  // addNode() will add a new node  // to the list function addNode(data) {     // Create a new node     let 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 function swap(n1, n2) {     let 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         let temp = node1.next;         node1.next = node2.next;         node2.next = temp;     }     else {         console.log("Swapping is not possible");     } }  // display() will display all the // nodes present in the list function display() {     // Node current will point to head     let current = head;      if (head == null) {         console.log("List is empty");         return;     }     while (current != null) {         // Prints each node by incrementing          // pointer         console.log(current.data);         current = current.next;     } }  // Add nodes to the list addNode(1); addNode(2); addNode(3); addNode(4); addNode(5); addNode(6); addNode(7);  console.log("Original list:"); display();  // Swaps node 2 with node 5 swap(6, 1);  console.log("List after swapping nodes:"); display(); // This code contributed by aashish1995

Output

Original list: 1 2 3 4 5 6 7 List after swapping nodes: 6 2 3 4 5 1 7

Complexity Analysis:

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

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