141. Linked List Cycle

class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        slow, fast = head, head
        
        while slow and fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            
            if slow == fast:
                return True
            
        return False

• use the classic fast and slow pointer algorithm to find the cycle in a linked-list.

Categories:: linked-list, slow-fast-pointer