1. 原始题目
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.
2. 题目理解
判断链表中是否有环。这个问题也是用快慢指针来解决。快指针每次走两步，慢指针每次走一步。如果存在环，那么快指针总会和慢指针在环中相遇！否则无环则快指针首先指向链表尾。关于快指针为何每次走两步，不能走3，4步的证明：知乎。
3. 解题
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 hasCycle(self, head): 9 """ 10 :type head: ListNode 11 :rtype: bool 12 """ 13 if not head: 14 return False 15 cur = head 16 pre = head 17 if not cur.next: 18 return False 19 while pre: # 因为pre总是在前面，所以判断pre即可 20 pre = pre.next 21 cur = cur.next 22 if pre: # 注意每次额外next操作时，都要先判断当前是否不为空结点 23 pre = pre.next 24 if pre==cur: 25 return True 26 return False