Description: Given head
, the head of a linked list, determine if the linked list has a cycle in it.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next
pointer. Internally, pos
is used to denote the index of the node that tail's next
pointer is connected to. Note that pos
is not passed as a parameter. Return true
if there is a cycle in the linked list. Otherwise, return false
.
Link:https://leetcode.com/problems/linked-list-cycle/
Examples:
Example 1: Input: head = [3,2,0,-4], pos = 1 Output: true Explanation: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed). Example 2: Input: head = [1,2], pos = 0 Output: true Explanation: There is a cycle in the linked list, where the tail connects to the 0th node. Example 3: Input: head = [1], pos = -1 Output: false Explanation: There is no cycle in the linked list.
思路: 没有环的链表终止条件是node.next = None, 有环的链表终止条件是node.next 在已经遍历过的nodes中出现过。
# Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def hasCycle(self, head): """ :type head: ListNode :rtype: bool """ if not head: return False if not head.next: return False p = head nodes = [] while p and (p not in nodes): nodes.append(p) p = p.next if not p: return False return True
日期: 2020-11-21 最近的实验被自己写出很多bug,从烦躁到越来越平静地debug,甚至很快乐