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Linked List Cycle II

Difficulty: Medium

Problem Description

Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null. 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 (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter. Do not modify the linked list.

Key Insights

  • A linked list can have a cycle if a node's next pointer points back to a previous node.
  • We can use Floyd's Tortoise and Hare algorithm, which uses two pointers moving at different speeds to detect cycles.
  • Once a cycle is detected, we can find the starting node of the cycle by moving one pointer to the head and keeping the other at the meeting point.

Space and Time Complexity

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

Solution

We will utilize Floyd's Tortoise and Hare algorithm to detect the cycle. The algorithm involves two pointers: a slow pointer that moves one step at a time and a fast pointer that moves two steps at a time. If there is a cycle, the fast pointer will eventually meet the slow pointer. Once a cycle is detected, we can find the starting point of the cycle by resetting one pointer to the head and moving both pointers one step at a time until they meet again.

Code Solutions

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def detectCycle(head: ListNode) -> ListNode:
    if not head or not head.next:
        return None

    slow = head
    fast = head

    # Phase 1: Detect if a cycle exists
    while fast and fast.next:
        slow = slow.next          # Move slow pointer by 1
        fast = fast.next.next     # Move fast pointer by 2
        if slow == fast:          # Cycle detected
            break
    else:
        return None                # No cycle

    # Phase 2: Find the entry point of the cycle
    slow = head
    while slow != fast:
        slow = slow.next
        fast = fast.next

    return slow                  # The start node of the cycle
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