Linked List Cycle II - Solution
Solutions and explanations

Video Explanation

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

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        visited = set()
        cur = head
        while cur:
            if cur in visited:
                return cur      # Cycle entry found
            visited.add(cur)
            cur = cur.next
        return None     # No Cycle 

Complexity Analysis

Here, n is the number of elements in the linked list.

  • Time Complexity: O(n)
  • Space Complexity: O(n)

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

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        # Step 1: Detect the cycle
        slow, fast = head, head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast: 
                break   # cycle detected
        else: # else runs only if the while loop ends naturally without 'break'
            return None # no cycle
        
        # Step 2: Find the start node of the cycle
        slow1 = head
        while slow != slow1:
            slow = slow.next
            slow1 = slow1.next
        return slow

Complexity Analysis

Here, n is the number of elements in the linked list.

  • Time Complexity: O(n)
  • Space Complexity: O(1)