Split a Circular Linked List into two halves
Original Linked List
Result Linked List 1
Result Linked List 2
If there are odd number of nodes, then first list should contain one extra.
Thanks to Geek4u for suggesting the 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.
Original Circular Linked List 11 2 56 12 First Circular Linked List 11 2 Second Circular Linked List 56 12
Time Complexity: O(n)
Please write comments if you find any bug in above code/algorithm, or find other ways to solve the same problem
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