How would you split a circular linked list into two equal halves?

• A. Use two pointers to locate midpoint and then carve the circle into two halves in a single pass.
• B. Break the list, compute halves in separate steps.
• C. Remove alternate nodes to form halves.
• D. Use slow/fast pointers to find the midpoint, then break the circle to form two circular lists representing the halves.

Answer: (D)

Locating the split point in a circular linked list without a null terminator and then forming two equal circular halves is best done with the slow/fast pointer technique and a precise break of the circle. Start with two pointers at the head: slow moves one step at a time and fast moves two steps. Because the list is circular, you stop when fast.next would loop back to the head (or when fast.next.next would), which places slow right at the end of the first half. The start of the second half is slow.next, so define head2 as that node. Break the first half by setting slow.next = head, making the first portion a closed circle from head to slow. Then close the second half by connecting its last node to head2. The last node in the second half is fast.next, so set fast.next.next = head2. You’ll end up with two circular lists: one from head to slow, and another from head2 to fast.next. This approach achieves the split in one pass and preserves the circular structure of both halves (assuming an even number of nodes; odd length would require a small adjustment for the middle node).