问题：
# 给定一个链表，判断链表中是否有环。
#
# 如果链表中有某个节点，可以通过连续跟踪 next 指针再次到达，则链表中存在环。 为了表示给定链表中的环，我们使用整数 pos 来表示链表尾连接到链表中的
# 位置（索引从 0 开始）。 如果 pos 是 -1，则在该链表中没有环。注意：pos 不作为参数进行传递，仅仅是为了标识链表的实际情况。
#
# 如果链表中存在环，则返回 true 。 否则，返回 false 。
方法一：暴力，hash/set
# leetcode submit region begin(Prohibit modification and deletion) # 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 """ res = set() while head and head.next: if head not in res: res.add(head) else: return True head = head.next return False # leetcode submit region end(Prohibit modification and deletion)
方法二：双指针法
# leetcode submit region begin(Prohibit modification and deletion) # 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 """ if not head or not head.next: return False slow = head fast = head.next while slow != fast: if not fast or not fast.next: return False slow = slow.next fast = fast.next.next return True # leetcode submit region end(Prohibit modification and deletion) # leetcode submit region begin(Prohibit modification and deletion) # 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 """ if not head or not head.next: return False slow = head fast = head.next while slow != fast: if not fast or not fast.next: return False slow = slow.next fast = fast.next.next return True # leetcode submit region end(Prohibit modification and deletion)
原文：https://www.cnblogs.com/demo-deng/p/14773733.html