Linked List Cycle

Leetcode #141

Problem

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.
 
Constraints:
The number of the nodes in the list is in the range [0, 104].
-105 ⇐ Node.val ⇐ 105
pos is -1 or a valid index in the linked-list.
 
Follow up: Can you solve it using O(1) (i.e. constant) memory?

My Solution: (Hash Set)

# 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:
        """
        - hash nodes we've seen
        - iterate till repeat node
        """
        seen = set()
        curr = head
        if not head:
            return False
 
        while curr.next:
            if curr in seen:
                return True
            else:
                seen.add(curr)
                curr = curr.next
        return False

the better solution is two pointer (fast and slow).

Optimal Solution: (Fast and Slow Pointers):

# 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:
        """
        - two pointers
            - fast moves by 2
            - slow moves by 1 
        """
        slow = fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                return True
        return False