Description
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.
Note: Do not modify the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.
Follow up:
Can you solve it without using extra space?
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。
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。
说明：不允许修改给定的链表。
示例 1：
输入：head = [3,2,0,-4], pos = 1
输出：tail connects to node index 1
解释：链表中有一个环，其尾部连接到第二个节点。
示例 2：
输入：head = [1,2], pos = 0
输出：tail connects to node index 0
解释：链表中有一个环，其尾部连接到第一个节点。
示例 3：
输入：head = [1], pos = -1
输出：no cycle
解释：链表中没有环。
进阶：
你是否可以不用额外空间解决此题？
思路
- 借助上一题的思路，我们使用两个指针，分别是slow和fast，slow指针每次向后走一个，fast指针每次向后走两步.
- 此题目借鉴了求链表中倒数第k个节点的思想.
- 基本思路是先求得环中一共有多少个节点m.
- 然后slow，fast重置为head,将fast往后移动m-个节点，然后slow和fast同时往后移动，当fast.next == slow时返回slow即可.
# -*- coding: utf-8 -*- # @Author: 何睿 # @Create Date: 2019-01-12 10:09:56 # @Last Modified by: 何睿 # @Last Modified time: 2019-01-12 12:05:39 # Definition for singly-linked list. class ListNode(object): def __init__(self, x): self.val = x self.next = None class Solution(object): def detectCycle(self, head): """ :type head: ListNode :rtype: ListNode """ if not head or not head.next: return None count = 1 slow, fast = head, head.next while fast and fast.next: # 找到第一次相遇的节点 if fast == slow: # 确定环中节点的个数 fast = fast.next while fast and fast.next: # 如果fast指针和slow指针相遇，说明环已经走完一趟 # 结束循环 if fast == slow: break # 否则count自增一次 count += 1 slow, fast = slow.next, fast.next.next # 重置slow和fast只想首节点 slow, fast = head, head # 将fast往后移count-1个（即环中节点个数少一个节点） for _ in range(count - 1): fast = fast.next while fast: # 当fast.next 指向slow时，说明已经到达结尾 if fast.next == slow: return slow fast = fast.next slow = slow.next return slow slow, fast = slow.next, fast.next.next # 如果遍历的所有节点fast和slow都不等，说明没有环 return None
源代码文件在这里.
