Skip to main content

Merge sort a linked list

Sorting a linked list is much more complex than sorting an array. Because you need to be wary of links at each step and update them.

The most convenient ways of sorting a linked list is insertion sort, where in you remove one node at a time from list and insert them to the sorted list in correct position.

But merge sorting a linked list uses another approach. It splits the list into two halves, sorts them recursively and then merges them maintaining ascending order of key values.

So the three basic parts of this approach are
  1. Splitting the list into two halves of equal length
  2. Sorting these halves
  3. Merging the halves
Step 2 is in fact not needed if the list has only one node. One node is sorted. Hence the merge sort algorithm is

  1. If the list has more than one node
  2. Splitting the list into two halves of equal length
  3. Sorting these halves recursively using steps 2 to 4
  4. Merging the halves 

 Split the list into halves

We have seen in this post how to find the mid point of a linked list. Let us use this approach.

We need to use two pointers say slow and fast. Slow moves one node at a time. And fast moves two nodes at a time. When fast has reached end of the list, slow has reached mid point. 

And also we should detach first half of the list from second half.

NODEPTR split_list(NODEPTR head)
{
    NODEPTR slow,fast;
    slow =head; 
    fast = head;
    if(head==NULL || head->next==NULL)     
       return head; /* if only one node is present, return*/
    NODEPTR temp;
    while(fast!=NULL)         
    {       
        fast = fast->next;        
        if(fast!=NULL){
           temp = slow;
           fast = fast->next;
           slow = slow->next;
        }           
     }     
     temp->next  = NULL;/*detach first half of list from second half*/
     return slow; /*this is the head of second half of list*/
}

 Merge sorted sublists

To merge the two sublists which are already sorted we need to compare nodes from two lists, insert the smaller one into the merged list and move to next node in that list. This process has to be repeated until all nodes of both lists are merged. If the two sublists are of unequal length then one list will have nodes remaining, which should be added to merged list

NODEPTR merge_sorted_lists(NODEPTR head1,NODEPTR head2)
{
 NODEPTR newlist = NULL;
 while(head1!=NULL && head2!=NULL )
 {
    if(head1->val <head2->val)
  {
  NODEPTR temp = head1;
  head1 = head1->next; 
  temp->next = NULL;
  newlist = append_node(newlist,temp);
  }
    else
  {
  NODEPTR temp = head2;
  head2 = head2->next; 
  temp->next = NULL;
  newlist = append_node(newlist,temp);
  }
 }
 while(head1!=NULL)
 {
  NODEPTR temp = head1;
  head1 = head1->next; 
  temp->next = NULL;
  newlist = append_node(newlist,temp);
  }
 while(head2!=NULL)
 {
  NODEPTR temp = head2;
  head2 = head2->next; 
  temp->next = NULL;
  newlist = append_node(newlist,temp);
 }
 return newlist;  
}
 

Merge sort the list

 Let us use these two functions in the algorithm given earlier to merge sort the list.

NODEPTR merge_sort(NODEPTR head)
{
     if(head==NULL || head->next==NULL)
       return head; /* list has one node or is empty*/
     NODEPTR mid = split_list(head);
     head = merge_sort(head);/* sort first sublist*/
     mid = merge_sort(mid);/* sort second sublist*/
     head = merge_sorted_lists(head,mid);/* merge these*/
     return head;
}

You can download the entire program from here 
Labels:

Comments

Post a Comment

Popular posts from this blog

Delete a node from doubly linked list

Deletion operation in DLL is simpler when compared to SLL. Because we don't have to go in search of previous node of to-be-deleted node.  Here is how you delete a node Link previous node of node of to-be-deleted to next node. Link next node of node of to-be-deleted to previous node. Free the memory of node of to-be-deleted Simple, isn't it. The code can go like this. prevnode = delnode->prev; nextnode = delnode->next; prevnode->next = nextnode; nextnode->prev = prevnode; free(delnode); And that is it. The node delnode is deleted. But we should always consider boundary conditions. What happens if we are trying to delete the first node or last node? If first node is to be deleted, its previous node is NULL. Hence step 3 should not be used.  And also, once head is deleted, nextnode becomes head . Similarly if last node is to be deleted, nextnode is NULL. Hence step 4 is as strict NO NO. And we should set prevnode to tail. After we put these things together, we have...
Post a Comment
Read more

Program to delete a node from linked list

How do you remove a node from a linked list? If you have to delete a node, first you need to search the node. Then you should find its previous node. Then you should link the previous node to the next node. If node containing 8 has to be deleted, then n1 should be pointing to n2. Looks quite simple. Isn't it? But there are at least two special cases you have to consider. Of course, when the node is not found. If the node is first node of the list viz head. If the node to be deleted is head node, then if you delete, the list would be lost. You should avoid that and make the second node as the head node. So it becomes mandatory that you return the changed head node from the function.   Now let us have a look at the code. #include<stdio.h> #include<stdlib.h> struct node { int data; struct node * next; }; typedef struct node * NODEPTR; NODEPTR create_node ( int value) { NODEPTR temp = (NODEPTR) malloc( size...
3 comments
Read more

Binary tree deletion

Do not get all scared and worried. It is not rocket science (or as I would like to call it - it is not regex). Remember deleting a node from linked list. When you deleted a node, you did the following 2 steps Free the memory of the node Link the previous node of the deleted node to the next node You will have to do these things in binary tree too. But the difficulty here is do you link the previous node - or parent node in tree terminology, to the left child of deleted node? Or to the right child? You can not link both children, as the parent already may have one child node. So before we solve this, let us categorize the deletion with the help of a diagram. Consider the cases The node to be deleted is a leaf node. That is, it does not have left or right child. In this case, link the parent to NULL in place of deleted node.  If you want to delete 13 - which is a leaf node, you set 14->left to NULL and free memory of node 13. If you want to delete 7 - which is also a leaf node and...
Post a Comment
Read more