Sort a linked list that is sorted alternating ascending and descending orders? - GeeksforGeeks
Given a Linked List. The Linked List is in alternating ascending and descending orders. Sort the list efficiently.
Example:
Read full article from Sort a linked list that is sorted alternating ascending and descending orders? - GeeksforGeeks
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->43->53->67->89->NULL
A Simple Solution is to use Merge Sort for linked List. This solution takes O(n Log n) time.
An Efficient Solution works in O(n) time. Below are all steps.
1. Separate two lists.
2. Reverse the one with descending order
3. Merge both lists.
1. Separate two lists.
2. Reverse the one with descending order
3. Merge both lists.
class LinkedList{ Node head; // head of list /* 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; } // 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; }}