Given a linked list, determine if it has a cycle in it.
用两个pointer，slow 和 fast
同时出发，slow每次往前一步，fast每次往前两步
如果相遇，说明有loop
否则没有
一定会相遇，因为在slow进入loop的时候，fast已经在loop里面，每次追一步，一定能追上（小学奥数追及问题）
def hasCycle(self, head):
"""
:type head: ListNode
:rtype: bool
"""
slow , fast = head , head
while fast and fast.next:
slow,fast = slow.next,fast.next.next
if slow is fast:
return True
return False
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使用快慢指针就好啦
本文标题：141. Linked List Cycle
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