142. Linked List Cycle II¶

#
# @lc app=leetcode id=142 lang=python3
#
# [142] Linked List Cycle II
#
# https://leetcode.com/problems/linked-list-cycle-ii/description/
#
# algorithms
# Medium (48.87%)
# Likes:    12740
# Dislikes: 891
# Total Accepted:    1.2M
# Total Submissions: 2.3M
# 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.

# from typing import Optional


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


class Solution:
    def detectCycle(self, head):

        # Solution by set:

        # v = set()

        # while head:

        #     if head in v:
        #         return head

        #     v.add(head)
        #     head = head.next

        # return

        # solution by fast&slow pointers

        f = s = head

        while True:

            for _ in range(2):
                if not f:
                    return None
                f = f.next

            s = s.next

            if f == s:
                break

        i = head
        j = s

        while i != j:
            i = i.next
            j = j.next

        return i

# @lc code=end