2674. Split a Circular Linked List
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)
:
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};
}Code Editor (Testing phase)
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