Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
思路：笨办法是每个节点再开辟一个属性存放是否访问过，这样遍历一遍即可知道是否有环。但为了不增加额外的空间，可以设置两个指针，一个一次走一步，另一个一次走两步，如果有环则两个指针一定会再次相遇，反之则不会。
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# @param head, a ListNode
# @return a boolea
def hasCycle(self, head):
if head is None:
return False
p1 = head
p2 = head
while True:
if p1.next is not None:
p1=p1.next.next
p2=p2.next
if p1 is None or p2 is None:
return False
elif p1 == p2:
return True
else:
return False
return False