Given a singly linked list, remove all the nodes which have a greater value on the right side.
Examples:
Input: 12->15->10->11->5->6->2->3->NULL
Output: 15->11->6->3->NULL
Explanation: 12, 10, 5 and 2 have been deleted because there is a
greater value on the right side. When we examine 12,
we see that after 12 there is one node with a value
greater than 12 (i.e. 15), so we delete 12. When we
examine 15, we find no node after 15 that has a value
greater than 15, so we keep this node. When we go like
this, we get 15->6->3
Input: 10->20->30->40->50->60->NULL
Output: 60->NULL
Explanation: 10, 20, 30, 40, and 50 have been deleted because
they all have a greater value on the right side.
Input: 60->50->40->30->20->10->NULL
Output: No Change.
Method 1 (Simple):
Use two loops. In the outer loop, pick nodes of the linked list one by one. In the inner loop, check if there exists a node whose value is greater than the picked node. If there exists a node whose value is greater, then delete the picked node.
Time Complexity: O(n^2)
Method 2 (Use Reverse):
Thanks to Paras for providing the below algorithm.
1. Reverse the list.
2. Traverse the reversed list. Keep max till now. If the next node is less than max, then delete the next node, otherwise max = next node.
3. Reverse the list again to retain the original order.
Time Complexity: O(n)
Thanks to R.Srinivasan for providing the code below.
Javascript
<script>// Javascript program to delete nodes which // have a greater value on right side // head of listvar head; // Linked list Node class Node { constructor(val) { this.data = val; this.next = null; }}/* Deletes nodes which have a node with greater value node on left side */function delLesserNodes() { // 1.Reverse the linked list reverseList(); /* 2. In the reversed list, delete nodes which have a node with greater value node on left side. Note that head node is never deleted because it is the leftmost node. */ _delLesserNodes(); /* 3. Reverse the linked list again to retain the original order */ reverseList();}/* Deletes nodes which have greater value node(s) on left side */function _delLesserNodes() { var current = head; // Initialise max var maxnode = head; var temp; while (current != null && current.next != null) { /* If current is smaller than max, then delete current */ if (current.next.data < maxnode.data) { temp = current.next; current.next = temp.next; temp = null; } /* If current is greater than max, then update max and move current */ else { current = current.next; maxnode = current; } }}// Utility functions // Inserts a new Node at front of the function push(new_data) { /* 1 & 2: Allocate the Node & Put in the data */ var new_node = new Node(new_data); // 3. Make next of new Node as head new_node.next = head; // 4. Move the head to point to // new Node head = new_node;}// Function to reverse the linked list function reverseList() { var current = head; var prev = null; var next; while (current != null) { next = current.next; current.next = prev; prev = current; current = next; } head = prev;}// Function to print linked list function printList() { var temp = head; while (temp != null) { document.write(temp.data + " "); temp = temp.next; } document.write();}// Driver code/* Constructed Linked List is 12->15->10->11-> 5->6->2->3 */push(3);push(2);push(6);push(5);push(11);push(10);push(15);push(12);document.write("Given Linked List<br/>");printList();delLesserNodes();document.write("<br/>Modified Linked List<br/>");printList();// This code is contributed by aashish1995 </script> |
Output:
Given Linked List 12 15 10 11 5 6 2 3 Modified Linked List 15 11 6 3
Time Complexity: O(n)
Auxiliary Space: O(1)
Method 3:
The other simpler method is to traverse the list from the start and delete the node when the current Node < next Node. To delete the current node, follow this approach.Let us assume you have to delete current node X:
- Copy next node’s data into X i.e X.data = X.next.data.
- Copy next node’s next address i.e X.next = X.next.next.
Move forward in the List only when the current Node is > the next Node.
Javascript
<script>// Javascript program for above approach// This class represents a single node // in a linked listclass Node { constructor(val) { this.data = val; this.next = null; }}// This is a utility class for linked list// This function creates a linked list from a// given array and returns headfunction createLL(arr) { var head = new Node(arr[0]); var temp = head; var newNode = null; for (i = 1; i < arr.length; i++) { newNode = new Node(arr[i]); temp.next = newNode; temp = temp.next; } return head;}// This function prints given linked listfunction printLL(head) { while (head != null) { document.write(head.data + " "); head = head.next; } document.write("<br/>");}// Main functionfunction deleteNodesOnRightSide(head) { if (head == null || head.next == null) return head; var nextNode = deleteNodesOnRightSide(head.next); if (nextNode.data > head.data) return nextNode; head.next = nextNode; return head;}var arr = [12, 15, 10, 11, 5, 6, 2, 3];var head = createLL(arr);document.write("Given Linked List<br/>");printLL(head);head = deleteNodesOnRightSide(head);document.write("<br/>Modified Linked List<br/>");printLL(head);// This code is contributed by aashish1995 </script> |
Output:
Given Linked List 12 15 10 11 5 6 2 3 Modified Linked List 15 11 6 3
Time Complexity: O(n)
Auxiliary Space: O(1)
Please refer complete article on Delete nodes which have a greater value on right side for more details!
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