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