【leetcode】142. 环形链表 II

叁歲伎倆 2022-03-16 13:11 212阅读 0赞

题目描述：  
给定一个链表，返回链表开始入环的第一个节点。 如果链表无环，则返回 null。

示例：

> 有环示例：  
> ![20190226214956725.png][]

链接：[https://leetcode-cn.com/problems/linked-list-cycle-ii/][https_leetcode-cn.com_problems_linked-list-cycle-ii]

--------------------

此题与【[141. 环形链表][141.]】无太大差别，重点关注第三种算法。

第一种：

class Solution(object):
        def detectCycle(self, head):
            """ :type head: ListNode :rtype: ListNode """
            record = set()
            while head:
                if head in record:
                    return head
                record.add(head)
                head = head.next
            return None

第二种：

class Solution(object):
        def detectCycle(self, head):
            """ :type head: ListNode :rtype: ListNode """
            while head:
                if not hasattr(head, "visited"):
                    head.visited = True
                    head = head.next
                else:
                    return head
            else:
                return None

第三种：

*  首先，通过快慢指针判断链表是否循环。也就是快慢指针是否会相遇；
 *  如果相遇，慢指针继续往后走；同时，在链表头处**再派出一个指针**，此指针速度也为1；
 *  当新指针与慢指针相遇时，相遇点为环点。

详情可见：[https://blog.csdn.net/xy010902100449/article/details/48995255\#commentsedit][https_blog.csdn.net_xy010902100449_article_details_48995255_commentsedit]

class Solution(object):
        def detectCycle(self, head):
            """ :type head: ListNode :rtype: ListNode """
            slow = fast = head
            while fast and fast.next:
                slow, fast = slow.next, fast.next.next
                if slow is fast:
                    break
            else:
                return None
    
            while head != slow:
                head = head.next
                slow = slow.next
            return head

[20190226214956725.png]: /images/20220316/38b0d55070274af79cdc34dbf9989bd4.png
[https_leetcode-cn.com_problems_linked-list-cycle-ii]: https://leetcode-cn.com/problems/linked-list-cycle-ii/
[141.]: https://blog.csdn.net/qq_41359051/article/details/87951745
[https_blog.csdn.net_xy010902100449_article_details_48995255_commentsedit]: https://blog.csdn.net/xy010902100449/article/details/48995255#commentsedit