Java Program For Sorting A Linked List That Is Sorted Alternating Ascending And Descending Orders Last Updated : 08 Dec, 2022 Summarize Comments Improve Suggest changes Share Like Article Like Report Given a Linked List. The Linked List is in alternating ascending and descending orders. Sort the list efficiently. Example: Input List: 10 -> 40 -> 53 -> 30 -> 67 -> 12 -> 89 -> NULL Output List: 10 -> 12 -> 30 -> 40 -> 53 -> 67 -> 89 -> NULL Input List: 1 -> 4 -> 3 -> 2 -> 5 -> NULL Output List: 1 -> 2 -> 3 -> 4 -> 5 -> NULLRecommended: Please solve it on "PRACTICE" first, before moving on to the solution. Simple Solution: Approach: The basic idea is to apply to merge sort on the linked list. The implementation is discussed in this article: Merge Sort for linked List.Complexity Analysis: Time Complexity: The merge sort of linked list takes O(n log n) time. In the merge sort tree, the height is log n. Sorting each level will take O(n) time. So time complexity is O(n log n).Auxiliary Space: O(n log n), In the merge sort tree the height is log n. Storing each level will take O(n) space. So space complexity is O(n log n). Efficient Solution: Approach: Separate two lists.Reverse the one in descending orderMerge both lists. Diagram: Below are the implementations of the above algorithm: Java // Java program to sort a linked list // that is alternatively sorted in // increasing and decreasing order import java.io.*; public class LinkedList { // head of list Node head; // Linked list Node class Node { int data; Node next; Node(int d) { data = d; next = null; } } Node newNode(int key) { return new Node(key); } /* This is the main function that sorts the linked list.*/ void sort() { /* Create 2 dummy nodes and initialise as heads of linked lists */ Node Ahead = new Node(0), Dhead = new Node(0); // Split the list into lists splitList(Ahead, Dhead); Ahead = Ahead.next; Dhead = Dhead.next; // Reverse the descending list Dhead = reverseList(Dhead); // Merge the 2 linked lists head = mergeList(Ahead, Dhead); } // Function to reverse the linked list Node reverseList(Node Dhead) { Node current = Dhead; Node prev = null; Node next; while (current != null) { next = current.next; current.next = prev; prev = current; current = next; } Dhead = prev; return Dhead; } // Function to print linked list void printList() { Node temp = head; while (temp != null) { System.out.print(temp.data + " "); temp = temp.next; } System.out.println(); } // A utility function to merge // two sorted linked lists Node mergeList(Node head1, Node head2) { // Base cases if (head1 == null) return head2; if (head2 == null) return head1; Node temp = null; if (head1.data < head2.data) { temp = head1; head1.next = mergeList(head1.next, head2); } else { temp = head2; head2.next = mergeList(head1, head2.next); } return temp; } // This function alternatively splits // a linked list with head as head into two: // For example, 10->20->30->15->40->7 is // splitted into 10->30->40 and 20->15->7 // "Ahead" is reference to head of ascending // linked list // "Dhead" is reference to head of descending // linked list void splitList(Node Ahead, Node Dhead) { Node ascn = Ahead; Node dscn = Dhead; Node curr = head; // Link alternate nodes while (curr != null) { // Link alternate nodes in // ascending order ascn.next = curr; ascn = ascn.next; curr = curr.next; if (curr != null) { dscn.next = curr; dscn = dscn.next; curr = curr.next; } } ascn.next = null; dscn.next = null; } // Driver code public static void main(String args[]) { LinkedList llist = new LinkedList(); llist.head = llist.newNode(10); llist.head.next = llist.newNode(40); llist.head.next.next = llist.newNode(53); llist.head.next.next.next = llist.newNode(30); llist.head.next.next.next.next = llist.newNode(67); llist.head.next.next.next.next.next = llist.newNode(12); llist.head.next.next.next.next.next.next = llist.newNode(89); System.out.println("Given linked list"); llist.printList(); llist.sort(); System.out.println("Sorted linked list"); llist.printList(); } } // This code is contributed by Rajat Mishra OutputGiven linked list 10 40 53 30 67 12 89 Sorted linked list 10 12 30 40 53 67 89 Complexity Analysis: Time Complexity: O(n). One traversal is needed to separate the list and reverse them. The merging of sorted lists takes O(n) time.Auxiliary Space: O(1). No extra space is required. Please refer complete article on Sort a linked list that is sorted alternating ascending and descending orders? for more details! Comment More infoAdvertise with us Next Article Java Program For Arranging Single Linked List In Alternate Odd and Even Nodes Order K kartik Follow Improve Article Tags : Linked List Sorting Java Programs DSA Amazon Merge Sort Linked-List-Sorting +3 More Practice Tags : AmazonLinked ListMerge SortSorting Similar Reads Java Program For Arranging Single Linked List In Alternate Odd and Even Nodes Order Given a singly linked list, rearrange the list so that even and odd nodes are alternate in the list.There are two possible forms of this rearrangement. If the first data is odd, then the second node must be even. The third node must be odd and so on. 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