142. Linked List Cycle II

Medium

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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:

• The number of the nodes in the list is in the range [0, 104].

• -105 <= Node.val <= 105

• pos is -1 or a valid index in the linked-list.

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

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
from collections import defaultdict
class Solution:
    # Follow Up
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if head is None or head.next is None:
            return None
        slow = head
        fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                break
        if slow != fast:
            return None
        slow = head
        while slow != fast:
            slow = slow.next
            fast = fast.next
        return slow
                
    
    def detectCycleOld(self, head: Optional[ListNode]) -> Optional[ListNode]:
        s = set()
        ptr = head
        while ptr and ptr not in s:
            s.add(ptr)
            ptr = ptr.next
        return None if ptr is None else ptr