Linked List Cycle 2

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list.

Solution

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def detectCycle(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head is None or head.next is None:
            return None
        slow = fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                break
        if slow == fast:
            slow = head
            while slow != fast:
                slow = slow.next
                fast = fast.next
            return slow
        return None

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {

        if(head == null || head.next == null)
            return null;

        ListNode slow = head, fast = head;

        while(fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;

            if(slow == fast)
                break;
        }

        if(slow == fast) {
            slow = head;

            while(slow != fast) {
                slow = slow.next;
                fast = fast.next;
            }

            return slow;
        }

        return null;
    }
}