标签：NPU turn mod def sel pytho dex out code
题目描述：
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?
代码实现：
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 try: 14 fast = head.next 15 slow = head 16 while fast is not slow: 17 fast = fast.next.next 18 slow = slow.next 19 except: 20 # if there is an exception, we reach the end and there is no cycle 21 return None 22 23 # since fast starts at head.next, we need to move slow one step forward 24 slow = slow.next 25 while head is not slow: 26 head = head.next 27 slow = slow.next 28 29 return head
标签：NPU turn mod def sel pytho dex out code
原文地址：https://www.cnblogs.com/tbgatgb/p/10992375.html
