# 2674. Split a Circular Linked List

## Description

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.

A circular linked list is a normal linked list with the only difference being that the last node's next node, is the first node.

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:

• The number of nodes in list is in the range [2, 105]
• 0 <= Node.val <= 109
• LastNode.next = FirstNode where LastNode is the last node of the list and FirstNode is the first one

## Solutions

• /**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode() {}
*     ListNode(int val) { this.val = val; }
*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
ListNode a = list, b = list;
while (b.next != list && b.next.next != list) {
a = a.next;
b = b.next.next;
}
if (b.next != list) {
b = b.next;
}
ListNode list2 = a.next;
b.next = list2;
a.next = list;
return new ListNode[] {list, list2};
}
}

• /**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode() : val(0), next(nullptr) {}
*     ListNode(int x) : val(x), next(nullptr) {}
*     ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* a = list;
ListNode* b = list;
while (b->next != list && b->next->next != list) {
a = a->next;
b = b->next->next;
}
if (b->next != list) {
b = b->next;
}
ListNode* list2 = a->next;
b->next = list2;
a->next = list;
return {list, list2};
}
};

• # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
self, list: Optional[ListNode]
) -> List[Optional[ListNode]]:
a = b = list
while b.next != list and b.next.next != list:
a = a.next
b = b.next.next
if b.next != list:
b = b.next
list2 = a.next
b.next = list2
a.next = list
return [list, list2]


• /**
* type ListNode struct {
*     Val int
*     Next *ListNode
* }
*/
a, b := list, list
for b.Next != list && b.Next.Next != list {
a = a.Next
b = b.Next.Next
}
if b.Next != list {
b = b.Next
}
list2 := a.Next
b.Next = list2
a.Next = list
return []*ListNode{list, list2}
}

• /**
* class ListNode {
*     val: number
*     next: ListNode | null
*     constructor(val?: number, next?: ListNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.next = (next===undefined ? null : next)
*     }
* }
*/

list: ListNode | null,
): Array<ListNode | null> {
let a = list;
let b = list;
while (b.next !== list && b.next.next !== list) {
a = a.next;
b = b.next.next;
}
if (b.next !== list) {
b = b.next;
}
const list2 = a.next;
b.next = list2;
a.next = list;
return [list, list2];
}