题目来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/linked-list-cycle-ii

目前考虑到两种解法，但都需要辅助空间， 第一种 O(n) 第二种 O(1)

将走过的节点都记录在字典中，通过查询字典的key值是否存在来确定是否有环
时间复杂度为 O(n) , 空间复杂度为 O(n)

代码如下：

# -*- coding: utf-8 -*-# @Author : xaohuihui# @Time : 19-12-9# @File : detect_cycled_ii.py# Software : study"""环形链表ii"""class ListNode: def __init__(self, x): self.x = x self.next = None# Number.1def has_cycle(head: ListNode) -> ListNode: result = None if head and head.next: set_node = set() while head: if head in set_node: result = head break set_node.add(head) head = head.next return resultif __name__ == '__main__': # head=[3,2,0,4] pos= 1 node1 = ListNode(3) node2 = ListNode(2) node3 = ListNode(0) node4 = ListNode(4) node1.next = node2 node2.next = node3 node3.next = node4 node4.next = node2 result_node = has_cycle(node1) if result_node: start = node1 i = 0 while start: if result_node == start: print(f"tail connects to node index {i}") break i += 1 start = start.next else: print("no cycle")

输出如下：

tail connects to node index 1

使用快慢指针，快指针每次走两步，慢指针每次走一步。
如果单链表中有环，快慢指针肯定会相遇，如图a.所示，在相遇后，将快指针指向开始位置，结束第一次循环。
第二次循环，将快指针变为没次走一步，慢指针每次走一步，如图b.所示，如果再次相遇，该点就为环点
时间复杂度为 O(n) , 空间复杂度为 O(1)
特别注意： 若环点就在起始节点，第一次快慢指针相遇一定在环点 ，则fast和slow此时都指向起始节点，则第二次循环不必执行，如图c.所示

图a.

图b.

图c.

代码如下：

# -*- coding: utf-8 -*-# @Author : xaohuihui# @Time : 19-12-9# @File : detect_cycled_ii.py# Software : study"""环形链表ii"""class ListNode: def __init__(self, x): self.x = x self.next = None# NUmber.2def has_cycle(head: ListNode) -> ListNode: result = None if head and head.next: fast = slow = head while fast and fast.next: fast = fast.next.next slow = slow.next if fast == slow: fast = head break else: return result while fast != slow: fast = fast.next slow = slow.next result = fast return resultif __name__ == '__main__': # head=[3,2,0,4] pos= 0 node1 = ListNode(3) node2 = ListNode(2) node3 = ListNode(0) node4 = ListNode(4) node1.next = node2 node2.next = node3 node3.next = node4 node4.next = node1 result_node = has_cycle(node1) if result_node: start = node1 i = 0 while start: if result_node == start: print(f"tail connects to node index {i}") break i += 1 start = start.next else: print("no cycle")

输出结果

tail connects to node index 0