LC141. Linked List Cycle

# 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:
        if not head: return False
 
        prev = None
        orig_head = head
        while head.next is not None:
            tmp = head.next
            
            if tmp == orig_head:
                return True
 
            # Reverse the direction as we traverse the nodes
            # If a cycle exists in original list, then we'll
            # eventually travel back to the original head
            head.next = prev
            prev = head
            head = tmp
        
        return False

Complexity

Time: $O(n)$ — We’d be travelling to and fro throughout the list in the worst case
Space: $O(1)$