# Question

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?

# Solution: Floyd’s Tortoise and Hare

To find whether there is a circle, we can use two pointers, hare and tortoise, one advances twice as fast as the other. There exists a cycle when tortoise and hare reaches the same point, or hare reaches the end of the linked list.

Two pointers, one at head and the other at the meeting point of hare and tortoise, will meet at the start of the cycle if they travel at the same speed.

See Leetcode solution for detailed explanation.

```
# 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
hare, tortoise = head, head
while True:
try:
hare = hare.next.next
except AttributeError:
return None
tortoise = tortoise.next
if hare == tortoise:
break
node1, node2 = head, hare
while node1 != node2:
node1 = node1.next
node2 = node2.next
return node1
```