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Java Program For Sorting Linked List Which Is Already Sorted On Absolute Values

  • Last Updated : 22 Mar, 2022

Given a linked list that is sorted based on absolute values. Sort the list based on actual values.
Examples: 

Input:  1 -> -10 
Output: -10 -> 1

Input: 1 -> -2 -> -3 -> 4 -> -5 
Output: -5 -> -3 -> -2 -> 1 -> 4 

Input: -5 -> -10 
Output: -10 -> -5

Input: 5 -> 10 
Output: 5 -> 10

Source : Amazon Interview

A simple solution is to traverse the linked list from beginning to end. For every visited node, check if it is out of order. If it is, remove it from its current position and insert it at the correct position. This is the implementation of insertion sort for linked list and the time complexity of this solution is O(n*n).
A better solution is to sort the linked list using merge sort. Time complexity of this solution is O(n Log n).
An efficient solution can work in O(n) time. An important observation is, all negative elements are present in reverse order. So we traverse the list, whenever we find an element that is out of order, we move it to the front of the linked list. 
Below is the implementation of the above idea.

ever we find an element that is out of order, we move it to the front of the linked list. 
Below is the implementation of the above idea. 

Java




// Java program to sort a linked list,
// already sorted by absolute values
class SortList
{
    // Head of list
    static Node head; 
   
    // Linked list Node
    static class Node
    {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }
     
    // To sort a linked list by actual values.
    // The list is assumed to be sorted by
    // absolute values.
    Node sortedList(Node head)
    {
        // Initialize previous and current
        // nodes
        Node prev = head;
        Node curr = head.next;
         
        // Traverse list
        while(curr != null)
        {
            // If curr is smaller than prev,
            // then it must be moved to head
            if(curr.data < prev.data)
            {
                // Detach curr from linked list
                prev.next = curr.next;
                 
                // Move current node to beginning
                curr.next = head;
                head = curr;
                 
                // Update current
                curr = prev;
            }
             
            // Nothing to do if current element
            // is at right place
            else
            prev = curr;
         
            // Move current
            curr = curr.next;
        }
        return head;
    }
     
    /* Inserts a new Node at front of
       the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node 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 print linked list
    void printList(Node head)
    {
        Node temp = head;
        while (temp != null)
        {
           System.out.print(temp.data + " ");
           temp = temp.next;
        
        System.out.println();
    }
     
    // Driver code
    public static void main(String args[])
    {
        SortList llist = new SortList();
          
        /* Constructed Linked List is
           1->2->3->4->5->6->
           7->8->8->9->null */
        llist.push(-5);
        llist.push(5);
        llist.push(4);
        llist.push(3);
        llist.push(-2);
        llist.push(1);
        llist.push(0);
          
        System.out.println("Original List :");
        llist.printList(llist.head);
          
        llist.head = llist.sortedList(head);
  
        System.out.println("Sorted list :");
        llist.printList(llist.head);
    }
}
 
// This code is contributed by Amit Khandelwal(Amit Khandelwal 1).

Output: 

Original list :
0 -> 1 -> -2 -> 3 -> 4 -> 5 -> -5
Sorted list :
-5 -> -2 -> 0 -> 1 -> 3 -> 4 -> 5

Time Complexity: O(N)

Auxiliary Space: O(1)

Please refer complete article on Sort linked list which is already sorted on absolute values for more details!


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