## Question1

Given `head`

, the head of a linked list, determine if the linked list has a cycle in it.

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. **Note that pos is not passed as a parameter**.

Return `true`

*if there is a cycle in the linked list*. Otherwise, return `false`

.

**Example 1:**

```
Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).
```

## Algorithm1

This is a classic leetcode linked list problem. To find if it's cyclic, we need two nodes, one move one step forward and the other move two steps. If they finally meet, there must be a cycle, otherwise, the fast will loop until it reach the tail of list.

## Code1

/** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public boolean hasCycle(ListNode head) { if (head == null || head.next == null) return false; ListNode fast = head; ListNode slow = head; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; if (slow == fast) { return true; } } return false; } }

## Question2

This question is like a follow up question that require you to find the meeting point.

## Algorithm2

## Code2

/** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode detectCycle(ListNode head) { if (head == null || head.next == null) { return null; } boolean cycle = false; ListNode fast = head; ListNode slow = head; while (fast != null && fast.next != null) { fast = fast.next.next; slow = slow.next; if (fast == slow) { cycle = true; break; } } if (!cycle) return null; slow = head; while (slow != fast) { slow = slow.next; fast = fast.next; } return slow; } }