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142.py
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142.py
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'''
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?
'''
# 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 not head:
return None
slow, fast = head, head.next
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