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

Solution 1: Fast and Slow Pointers

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 $O(1)$.

Python3JavaC++GoTypeScript

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def splitCircularLinkedList(
        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]

/**
 * Definition for singly-linked list.
 * 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 {
    public ListNode[] splitCircularLinkedList(ListNode list) {
        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};
    }
}

/**
 * Definition for singly-linked list.
 * 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:
    vector<ListNode*> splitCircularLinkedList(ListNode* list) {
        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.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func splitCircularLinkedList(list *ListNode) []*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}
}

/**
 * Definition for singly-linked list.
 * 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)
 *     }
 * }
 */

function splitCircularLinkedList(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];
}

2674. Split a Circular Linked List 🔒

Description

Solutions

Solution 1: Fast and Slow Pointers

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