#
# @lc app=leetcode id=142 lang=python3
#
# [142] Linked List Cycle II
#
# https://leetcode.com/problems/linked-list-cycle-ii/description/
#
# algorithms
# Medium (40.76%)
# Likes:    5813
# Dislikes: 395
# Total Accepted:    566.8K
# Total Submissions: 1.3M
# Testcase Example:  '[3,2,0,-4]\n1'
#
# 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, 10^4].
# -10^5 <= Node.val <= 10^5
# pos is -1 or a valid index in the linked-list.
# 
# 
# 
# Follow up: Can you solve it using O(1) (i.e. constant) memory?
# 
#

# @lc code=start
# 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]:
        if not head:
            return None
        
        slow, fast = head, head.next
        while slow != fast:
            slow = slow.next if slow else None
            fast = fast.next.next if fast and fast.next else None

        if not slow:
            return None

        a = head
        b = slow.next
        while a != b:
            a = a.next
            b = b.next
        return a
            
# @lc code=end


#
# @lc app=leetcode id=142 lang=python3
#
# [142] Linked List Cycle II
#
# https://leetcode.com/problems/linked-list-cycle-ii/description/
#
# algorithms
# Medium (40.76%)
# Likes:    5813
# Dislikes: 395
# Total Accepted:    566.8K
# Total Submissions: 1.3M
# Testcase Example:  '[3,2,0,-4]\n1'
#
# 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, 10^4].
# -10^5 <= Node.val <= 10^5
# pos is -1 or a valid index in the linked-list.
# 
# 
# 
# Follow up: Can you solve it using O(1) (i.e. constant) memory?
# 
#

# @lc code=start
# 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]:
        if not head:
            return None
        
        slow, fast = head, head.next
        while slow != fast:
            slow = slow.next if slow else None
            fast = fast.next.next if fast and fast.next else None

        if not slow:
            return None

        a = head
        b = slow.next
        while a != b:
            a = a.next
            b = b.next
        return a
            
# @lc code=end


# # @lc app=leetcode id=142 lang=python3 # # [142] Linked List Cycle II # # https://leetcode.com/problems/linked-list-cycle-ii/description/ # # algorithms # Medium (40.76%) # Likes:    5813 # Dislikes: 395 # Total Accepted:    566.8K # Total Submissions: 1.3M # Testcase Example:  '[3,2,0,-4]\n1' # # 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, 10^4]. # -10^5 <= Node.val <= 10^5 # pos is -1 or a valid index in the linked-list. # # # # Follow up: Can you solve it using O(1) (i.e. constant) memory? # # # @lc code=start # Definition for singly-linked list. # class ListNode: #     def __init__(self, x): #         self.val = x #         self.next = None # @lc code=end