LeetCode 141. 环形链表 Linked List Cycle
Table of Contents
一、中文版
给定一个链表,判断链表中是否有环。
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。
示例 1:
输入:head = [3,2,0,-4], pos = 1
输出:true
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:true
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:false
解释:链表中没有环。
进阶:
你能用 O(1)(即,常量)内存解决此问题吗?
二、英文版
Given a linked list, determine if it has a cycle in it. 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. Example 1: Input: head = [3,2,0,-4], pos = 1 Output: true 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: true Explanation: There is a cycle in the linked list, where tail connects to the first node. Example 3: Input: head = [1], pos = -1 Output: false Explanation: There is no cycle in the linked list. Follow up: Can you solve it using O(1) (i.e. constant) memory? 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/linked-list-cycle 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
三、My answer
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def hasCycle(self, head: ListNode) -> bool: if not head: return False slow = head fast = head while fast != None and fast.next != None: fast = fast.next.next slow = slow.next if slow == fast: return True return False
四、解题报告
快慢指针。
快指针一次走两步,慢指针一次走一步。
如果快慢指针会相遇,则说明有环,否则无环。