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
<script>// JavaScript program to swap two // given nodes of a linked listclass Node { constructor(val) { this.data = val; this.next = null; }}// Head of listvar 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) var prevX = null, currX = head; while (currX != null && currX.data != x) { prevX = currX; currX = currX.next; } // Search for y (keep track of // prevY and currY) var 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 var temp = currX.next; currX.next = currY.next; currY.next = temp;}// Function to add Node at beginning // of listfunction push(new_data) { // 1. alloc the Node and put the data var 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 Nodefunction printList() { var tNode = head; while (tNode != null) { document.write(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);document.write( "Linked list before calling swapNodes()<br/> ");printList();swapNodes(4, 3);document.write( "<br/> Linked list after calling swapNodes() <br/>");printList();// This code is contributed by todaysgaurav</script> |
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:
Javascript
<script>// Javascript program to swap two given // nodes of a linked list// Represent a node of the singly // linked listclass Node { constructor(val) { this.data = val; this.next = null; }}// Represent the head and tail of // the singly linked listvar head = null;var tail = null;// addNode() will add a new node // to the listfunction addNode(data){ // Create a new node var 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 nodesfunction swap(n1 , n2) { var 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 var temp = node1.next; node1.next = node2.next; node2.next = temp; } else { document.write("Swapping is not possible"); }}// display() will display all the// nodes present in the listfunction display() { // Node current will point to head var current = head; if (head == null) { document.write("List is empty"); return; } while (current != null) { // Prints each node by incrementing // pointer document.write(current.data + " "); current = current.next; } document.write();}// Add nodes to the listaddNode(1);addNode(2);addNode(3);addNode(4);addNode(5);addNode(6);addNode(7);document.write("Original list:<br/> ");display();// Swaps node 2 with node 5swap(6, 1);document.write( "<br/>List after swapping nodes: <br/>");display();// This code contributed by aashish1995</script> |
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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