Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5. Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

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.

Note: 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.

Follow up: Can you solve it without using extra space?