Another Casual Coder

Deceiving problem: just need to split a circular linked-list into two. Problem is that we first need to count how many elements are there, and there may be duplicates. Easy solution is to change the first element to something unique, like -1. Then it becomes straightforward. Code is below, cheers, ACC.

Split a Circular Linked List - LeetCode

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.

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.

public ListNode[] SplitCircularLinkedList(ListNode list)
{
 ListNode[] retVal = new ListNode[2];
 if (list == null) return retVal;
 int tempFirst = list.val;
 list.val = -1;
 int count = 0;
 bool first = true;
 for (ListNode temp = list; first || temp.val != list.val; temp = temp.next)
 {
 first = false;
 count++;
 }
 retVal[0] = null;
 ListNode t0 = null;
 retVal[1] = null;
 ListNode t1 = null;
 for (int i = 0; i < count; i++)
 {
 if ((count % 2 == 0 && i < count / 2) || (count % 2 == 1 && i <= count / 2))
 {
 if (retVal[0] == null)
 {
 retVal[0] = new ListNode(list.val);
 t0 = retVal[0];
 }
 else
 {
 t0.next = new ListNode(list.val);
 t0 = t0.next;
 }
 }
 else
 {
 if (retVal[1] == null)
 {
 retVal[1] = new ListNode(list.val);
 t1 = retVal[1];
 }
 else
 {
 t1.next = new ListNode(list.val);
 t1 = t1.next;
 }
 }
 list = list.next;
 }
 t0.next = retVal[0];
 t1.next = retVal[1];
 retVal[0].val = tempFirst;
 return retVal;
}