# 142. Linked List Cycle II

Reference: LeetCode EPI 7.3

Difficulty: Medium

My Post: Potential Methods & Analysis (When you understand the process, coding is super easy)

## Problem

Given a linked list, return the node where the cycle begins. If there is no cycle, return

`null`

.

To represent a cycle in the given linked list, we use an integer

`pos`

which represents the position (0-indexed) in the linked list where tail connects to. If`pos`

is`-1`

, then there is no cycle in the linked list.

**Example:**

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

**Note:** Do not modify the linked list.

**Follow up:** Cau you solve it without using extra space? $O(1)$

## Analysis

### Hash Set

**Time:** $O(N)$**Space:** $O(N)$ ❌

### Mark Node

It modifies the linked list. ❌

**Time:** $O(N)$**Space:** $O(1)$

### Two Pointers

Reference: link

**Why could slow and fast meet?** See my analysis in 141. Linked List Cycle

**Notations:**

`head`

: the head node of the list.`entry`

: the entry node of the cycle.`meeting`

: the intersected node at which`slow`

and`fast`

meet.`L1`

: the distance between`head`

and`entry`

.`L2`

: the distance between`entry`

and`meeting`

.`C`

: the length of the cycle.`n`

: how many times`fast`

loops in the cycle.

**Floyd’s Cycle Detection:**

The first half is the same as in the two-pointer solution in `Problem I`

. To find the entry point, we can start traversing from `head`

and `slow`

at the meeting point in tandem. The node they meet is the entry node.

**Why is the node they meet the entry node?**

When `slow`

and `fast`

meet,

`slow`

moves`L1 + L2`

. (`head`

->`meeting`

)`fast`

moves`L1 + L2 + n * C`

.

Since each time `fast`

moves twice the step as `slow`

does, when they meet at `meeting`

node:

- We have:
`2 * Dis(slow) = 2 * (L1 + L2) = L1 + L2 + n * C = Dis(fast)`

=>`L1 + L2 = n * C`

. - Equivalent to:
`L1 = (n - 1) * C + (C - L2)`

where`C - L2`

is the distance between`meeting`

and`entry`

. - The equation indicates that if
`slow`

or`fast`

now starts moving from`meeting`

and a new node starts moving from`head`

, they will finally meet at the entry node!- Consider a very big
`L1`

and a relatively short`C`

. - Note that
`(n - 1) * C`

means that`slow`

or`fast`

moves from`meeting`

by`n - 1`

cycles and still stops at`meeting`

.

- Consider a very big

**Note:**

- It’d better to start
`slow`

and`fast`

from the head both. - Move first! Otherwise, they must meet at the beginning.

1 | public ListNode useTwoPointers(ListNode head) { |

**Time:** $O(L1 + L2 + L1) < O(N + N + N) = O(3N) = O(N)$

- Distance that
`slow`

moves before they meet: $O(L1 + L2)$ - Distance that the
`new node`

moves after`slow`

and`fast`

meet: $O(L1)$

**Space:** $O(1)$