leetcode-141 Linked List Cycle

#
# 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.
# 
# 
# 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.
# 

# Follow up: Can you solve it using O(1) (i.e. constant) memory?

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        fast = slow = head

        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                return True
        return False