Linked List Cycle II

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.

Follow up: Can you solve it without using extra space?

URL: https://leetcode.com/problems/linked-list-cycle-ii/

# 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 == None:
            return head
        else:
            fast = head
            slow = head

            has_cycle = False
            while fast != None and fast.next != None:
                slow = slow.next
                fast = fast.next.next
                if fast == slow:
                    has_cycle = True
                    break

            if has_cycle == False:
                return None

            slow = head
            while fast != slow:
                fast = fast.next
                slow = slow.next

            return slow