标签：nts 快慢指针 mod for http init lis linked 存在
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.
思路
这道题的主要思路就是我们先判断链表是否存在环(快慢指针快指针每次移动两步，慢指针每次移动一步，如果存在环的话，两个指针会重合)，然后找出这个环一共有几个节点(从重合的节点开始遍历一圈得到环中的节点数)，最后从头开始设置快慢指针，快指针先移动环的节点数步，然后快慢指针一起移动。当快慢指针重合时，指向的节点就表示环的入口节点。时间复杂度未O(n)， 空间复杂度为O(1)。
解决代码
1 # Definition for singly-linked list.
2 # class ListNode(object):
3 # def __init__(self, x):
4 # self.val = x
5 # self.next = None
6
7 class Solution(object):
8 def detectCycle(self, head):
9 """
10 :type head: ListNode
11 :rtype: ListNode
12 """
13 if not head or not head.next:
14 return None
15 one, two = head, head.next # 设置快慢指针
16 while one is not two: # 判断是否存在环
17 if not two or not two.next:
18 return None
19 one = one.next
20 two = two.next.next
21
22 count = 1 # 记录环中节点个数
23 while one is not two.next:
24 two = two.next
25 count += 1
26
27 one, two = head, head # 重新指向头节点
28 while count > 0: # two指针先移动count次
29 two = two.next
30 count -= 1
31 while one is not two: # 然后one，two指针一起移动。
32 one = one.next
33 two = two.next
34 return one # 返回one指向的节点就为环的入口节点。
【LeetCode每天一题】Linked List Cycle II(循环链表II)
标签：nts 快慢指针 mod for http init lis linked 存在
原文地址：https://www.cnblogs.com/GoodRnne/p/10957057.html
