Original Linked List

Result Linked List 1

Result Linked List 2

Algorithm:

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.

- C

` ` |

`/* 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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