[leetcode 141] Linked List Cycle

我们可以.用两个指针：one slow and one fast来解决这个问题，slow走一步，fast走两步，slow与fast相遇的地方是slow第一次到达，fast第二次到达的地方

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def hasCycle(self, head):        """        :type head: ListNode        :rtype: bool        """        slow=fast=head        while fast and fast.next:            slow=slow.next            fast=fast.next.next            if id(slow)==id(fast):                return True        return False