判断链表是否有环

解题思路一：快慢指针，快指针每次走两步，慢指针每次走一步，如果链表有环，则快慢指针就会相遇，否则，就是无环

class Solution:
    def hasCycle(self , head: ListNode) -> bool:
        if head is None:
            return False
        fast, slow = head, head
        while (fast != None and fast.next != None):
            fast = fast.next.next
            slow = slow.next
            if fast == slow:
                return True
        return False

思路二：利用 set （集合）是一个不重复的序列，遍历列表，如果存在之前遍历过的节点，就说明链表有环

class Solution：
	def hasCycle(self, head:ListNode) -> bool:
	if not head:
		return False
	dic = set()
	while head :
		if head in dic:
			return True
		dic.add(head)
		head = head.next
	return False