## Description

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 = , 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?

## Solutions

### 1. Two Pointers

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
# Time: O(n)
# Space: O(1)
class Solution:
def hasCycle(self, head: ListNode) -> bool:
if not head or not head.next:
return False
faster, slower = head, head
while faster and slower:
if faster.next:
faster = faster.next.next
else:
faster = None
slower = slower.next
if faster == slower:
return True
return False
# 17/17 cases passed (48 ms)
# Your runtime beats 89.92 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (16 MB)