Given a circular linked list list of positive integers, your task is to split it into 2 circular linked lists so that the first one contains the first half of the nodes in list (exactly ceil(list.length / 2) nodes) in the same order they appeared in list, and the second one contains the rest of the nodes in list in the same order they appeared in list.
Return an array answer of length 2 in which the first element is a circular linked list representing the first half and the second element is a circular linked list representing the second half.
Example 1:
Input: nums = [1,5,7] Output: [[1,5],[7]] Explanation: The initial list has 3 nodes so the first half would be the first 2 elements since ceil(3 / 2) = 2 and the rest which is 1 node is in the second half.
Example 2:
Input: nums = [2,6,1,5] Output: [[2,6],[1,5]] Explanation: The initial list has 4 nodes so the first half would be the first 2 elements since ceil(4 / 2) = 2 and the rest which is 2 nodes are in the second half.
Constraints:
list is in the range [2, 105]0 <= Node.val <= 109LastNode.next = FirstNode where LastNode is the last node of the list and FirstNode is the first oneProblem Overview: You receive the head of a circular linked list and must split it into two separate circular linked lists. The first half should contain the extra node when the length is odd. Both resulting lists must remain circular and preserve the original order of nodes.
Approach 1: Count Nodes Then Split (O(n) time, O(1) space)
The direct method first determines the total number of nodes in the circular list. Start at head and iterate until you reach the head again, counting nodes along the way. Once the length n is known, compute the midpoint (n + 1) / 2 so the first list gets the extra node when the size is odd.
Traverse again until you reach the midpoint node. Break the list by updating pointers: the midpoint becomes the tail of the first circular list, and the last node becomes the tail of the second circular list. Each tail must point back to its respective head. This approach is easy to reason about and useful when you want explicit control over positions in a linked list, but it requires two full traversals.
Approach 2: Fast and Slow Pointers (O(n) time, O(1) space)
The optimal method uses the classic two pointers technique. Initialize slow and fast at the head. Move slow one step and fast two steps at a time. Stop when fast reaches the end of the circular traversal (either fast.next == head or fast.next.next == head).
At that moment, slow points to the midpoint of the list. The node after slow becomes the head of the second circular list. Adjust pointers carefully: set slow.next to the original head to close the first circle, and connect the last node (tracked by fast) to the second head to close the second circle.
This works because the fast pointer covers twice the distance of the slow pointer, naturally landing the slow pointer at the midpoint. The algorithm performs only one traversal and does not require additional memory, making it the preferred solution for most linked list interview problems.
Recommended for interviews: The fast and slow pointer approach is what interviewers expect. It demonstrates mastery of pointer movement patterns and efficient traversal of circular structures. The counting approach still shows you understand pointer manipulation, but the two-pointer method highlights stronger algorithmic thinking and optimal single-pass design.
We define two pointers a and b, both initially pointing to the head of the linked list. Each iteration, pointer a moves forward one step, and pointer b moves forward two steps, until pointer b reaches the end of the linked list. At this point, pointer a points to half of the linked list nodes, and we break the linked list from pointer a, thus obtaining the head nodes of the two linked lists.
The time complexity is O(n), where n is the length of the linked list. It requires one traversal of the linked list. The space complexity is O(1).
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| Approach | Time | Space | When to Use |
|---|---|---|---|
| Count Nodes Then Split | O(n) | O(1) | When clarity is preferred and two passes over the list are acceptable |
| Fast and Slow Pointers | O(n) | O(1) | Preferred interview solution; single traversal using two-pointer technique |
Leetcode 2674 Split a Circular Linked List • AlgorithmicIQ • 22 views views
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