#141. 环形链表

题目：

思路：快慢指针问题

代码：

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

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        fast = head
        slow = head
        while(fast!=None and fast.next!=None): #判断如果这个链表至少有两个节点，才可以形成环
            fast = fast.next.next#快指针走一步
            slow =slow.next#慢指针走一步

            if(fast==slow):#如果他们相交，就返回True
                return True
        return False