Wednesday, 19 October 2016

Split a Circular Linked List into two halves

Split a Circular Linked List into two halves


                         Original Linked List  
                         Result Linked List 1  
                         Result Linked List 2  

1) Store the mid and last pointers of the circular linked list using tortoise and hare algorithm.

2) Make the second half circular.
3) Make the first half circular.
4) Set head (or start) pointers of the two linked lists.
In the below implementation, if there are odd nodes in the given circular linked list then the first result list has 1 more node than the second result list.
/* Program to split a circular linked list into two halves */
#include<stdio.h>
#include<stdlib.h>
 
/* structure for a node */
struct node
{
  int data;
  struct node *next;
};
 
/* Function to split a list (starting with head) into two lists.
   head1_ref and head2_ref are references to head nodes of
    the two resultant linked lists */
void splitList(struct node *head, struct node **head1_ref,
                                            struct node **head2_ref)
{
  struct node *slow_ptr = head;
  struct node *fast_ptr = head;
 
  if(head == NULL)
    return;
  
  /* If there are odd nodes in the circular list then
     fast_ptr->next becomes head and for even nodes
     fast_ptr->next->next becomes head */
  while(fast_ptr->next != head &&
         fast_ptr->next->next != head)
  {
     fast_ptr = fast_ptr->next->next;
     slow_ptr = slow_ptr->next;
  
 
 /* If there are even elements in list then move fast_ptr */
  if(fast_ptr->next->next == head)
    fast_ptr = fast_ptr->next;     
   
  /* Set the head pointer of first half */
  *head1_ref = head;   
      
  /* Set the head pointer of second half */
  if(head->next != head)
    *head2_ref = slow_ptr->next;
   
  /* Make second half circular */  
  fast_ptr->next = slow_ptr->next;
   
  /* Make first half circular */  
  slow_ptr->next = head;      
}
 
/* UTILITY FUNCTIONS */
/* Function to insert a node at the begining of a Circular
   linked lsit */
void push(struct node **head_ref, int data)
{
  struct node *ptr1 = (struct node *)malloc(sizeof(struct node));
  struct node *temp = *head_ref;
  ptr1->data = data; 
  ptr1->next = *head_ref;
   
  /* If linked list is not NULL then set the next of
    last node */
  if(*head_ref != NULL)
  {
    while(temp->next != *head_ref)
      temp = temp->next;       
    temp->next = ptr1;
  }
  else
     ptr1->next = ptr1; /*For the first node */
 
  *head_ref = ptr1;    
}
 
/* Function to print nodes in a given Circular linked list */
void printList(struct node *head)
{
  struct node *temp = head;
  if(head != NULL)
  {
    printf("\n");
    do {
      printf("%d ", temp->data);
      temp = temp->next;
    } while(temp != head);
  }
}
 
/* Driver program to test above functions */
int main()
{
  int list_size, i;
   
  /* Initialize lists as empty */
  struct node *head = NULL;
  struct node *head1 = NULL;
  struct node *head2 = NULL; 
 
  /* Created linked list will be 12->56->2->11 */
  push(&head, 12);
  push(&head, 56);  
  push(&head, 2);  
  push(&head, 11);  
 
  printf("Original Circular Linked List");
  printList(head);     
  
  /* Split the list */
  splitList(head, &head1, &head2);
  
  printf("\nFirst Circular Linked List");
  printList(head1); 
 
  printf("\nSecond Circular Linked List");
  printList(head2); 
   
  getchar();
  return 0;
}
Output:
Original Circular Linked List
11 2 56 12 
First Circular Linked List
11 2 
Second Circular Linked List
56 12 
Time Complexity: O(n)

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