Linked List Cycle II LeetCode Solution

Problem – Linked List Cycle II

Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail’s next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

Do not modify the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

Constraints:

• The number of the nodes in the list is in the range [0, 104].
• -105 <= Node.val <= 105
• pos is -1 or a valid index in the linked-list.

Linked List Cycle II LeetCode Solution in Python

class Solution(object):
    def detectCycle(self, head):
        slow = fast = head
        while fast and fast.next:
            slow, fast = slow.next, fast.next.next
            if slow == fast: break
        else: return None  # if not (fast and fast.next): return None
        while head != slow:
            head, slow = head.next, slow.next
        return head

Linked List Cycle II LeetCode Solution in C++

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode *slow = head, *fast = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast) break;
        }
        if (!(fast && fast->next)) return NULL;
        while (head != slow) {
            head = head->next;
            slow = slow->next;
        }
        return head;
    }
};

Linked List Cycle II LeetCode Solution in Java

public class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode slow = head, fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) break;
        }
        if (fast == null || fast.next == null) return null;
        while (head != slow) {
            head = head.next;
            slow = slow.next;
        }
        return head;
    }
}

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