LeetCode 2674: Split a Circular Linked List Solution

Master LeetCode problem 2674 (Split a Circular Linked List), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2674. Split a Circular Linked List

Medium

Linked List Two Pointers

Problem Explanation

Explanation:

The problem asks us to split a circular linked list into two separate circular linked lists. We need to find the two halve points and then rewire the pointers accordingly.

Algorithm:

Find the middle of the circular linked list using the slow and fast pointer technique.
Break the circular link by setting the next pointer of the middle node to null.
Set the head of the second half to the node next to the middle node.
Update the next pointer of the last node of the first half to point to the head of the first half.
Update the next pointer of the last node of the second half to point to the head of the second half.

Time Complexity: O(n) where n is the number of nodes in the circular linked list.

Space Complexity: O(1)

Solution Code

class ListNode {
    int val;
    ListNode next;
    
    public ListNode(int val) {
        this.val = val;
    }
}

public ListNode[] splitList(ListNode head) {
    if (head == null || head.next == head) {
        return new ListNode[] {head, null};
    }
    
    ListNode slow = head;
    ListNode fast = head;
    
    while (fast.next != head && fast.next.next != head) {
        slow = slow.next;
        fast = fast.next.next;
    }
    
    ListNode head2 = slow.next;
    slow.next = null;
    
    ListNode curr = head2;
    while (curr.next != head) {
        curr = curr.next;
    }
    
    curr.next = head2;
    
    return new ListNode[] {head, head2};
}

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