Problem URL: Split a circular linked list

We can use a slow and fast pointer to find the middle of the linked list. Then, we can set the next pointer of the last node to None to split the linked list into two. We can then return the head of the second linked list.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def splitCircularLinkedList(self, head: Optional[ListNode]) -> List[Optional[ListNode]]:
        slow, fast = head, head.next
        while fast.next != head:
            slow = slow.next

            if fast.next.next != head: 
                fast = fast.next.next
            else: 
                fast = fast.next

        next_list = slow.next
        slow.next = head
        fast.next = next_list
        return [head, next_list]

Time complexity: O(n) where n is the number of nodes in the linked list.
Space complexity: O(1)