LeetCode-in-All

142. Linked List Cycle II

Medium

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.

Example 1:

Input: head = [3,2,0,-4], pos = 1

Output: tail connects to node index 1

Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0

Output: tail connects to node index 0

Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1

Output: no cycle

Explanation: There is no cycle in the linked list.

Constraints:

Follow up: Can you solve it using O(1) (i.e. constant) memory?

Solution

# Definition for singly-linked list.
# class ListNode
#     attr_accessor :val, :next
#     def initialize(val)
#         @val = val
#         @next = nil
#     end
# end

# @param {ListNode} head
# @return {ListNode}
def detectCycle(head)
  return nil if head.nil? || head.next.nil?

  slow = head
  fast = head

  while fast && fast.next
    fast = fast.next.next
    slow = slow.next

    # Intersected inside the loop.
    break if slow == fast
  end

  return nil if fast.nil? || fast.next.nil?

  slow = head
  while slow != fast
    slow = slow.next
    fast = fast.next
  end

  slow
end