边工作边刷题：70天一遍leetcode: day 10

Linked List Cycle I/II

fast/slow指针的移动方式：fast先移动，然后检查fast again，同时移动fast和slow

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        fast = slow = head
        while fast:
            fast=fast.next
            if fast:
                fast=fast.next
                slow=slow.next
            
            if fast==slow:
                return True
        return False