Linked List 3 - Fast Slow Pointers
求交点:
Given a list, rotate the list to the right by k places, where k is non-negative.
Given 1->2->3->4->5 and k = 2, return 4->5->1->2->3.
3->5->1 is a cyclic list, so 3 is next node of 1.
3->5->1 is same with 5->1->3
insert a value 4: Return 5->1->3->4