Java Program For Sorting Linked List Which Is Already Sorted On Absolute Values
Last Updated :
09 Dec, 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
class SortList
{
static Node head;
static class Node
{
int data;
Node next;
Node( int d)
{
data = d;
next = null ;
}
}
Node sortedList(Node head)
{
Node prev = head;
Node curr = head.next;
while (curr != null )
{
if (curr.data < prev.data)
{
prev.next = curr.next;
curr.next = head;
head = curr;
curr = prev;
}
else
prev = curr;
curr = curr.next;
}
return head;
}
public void push( int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
void printList(Node head)
{
Node temp = head;
while (temp != null )
{
System.out.print(temp.data + " " );
temp = temp.next;
}
System.out.println();
}
public static void main(String args[])
{
SortList llist = new SortList();
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);
}
}
|
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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