CTCI 2.6

/*We could use a fast pointer and a slow pointer to find the node where the two pointer meets, since the 
list is circlelist, we are sure to find such a node. We could have:
Fast_length = Slow_length + n * circle_length, where n >= 1, since fast length must catch up slow one before 
it finishes one circle. We define the distance of list head to where circle start as K, and the start to where
two pointers meet as L, then it is easy to show: 
Slow_length = K + L, we restrict the speed of fast as two times of the slow, 2 nodes for fast and 
one node for slow,  then we could have
Slow_length = n * circle_length, then we have
K + L = n * circle_length, rewrite it as
K  = (n-1)*circle_length + (circle_length - L). From this, we know that from the meet point to the start point,
it is just the disctance from head to start point, plus some circle_length. Then we could have two slow
pointer start from head and meet point, where they meet is just the start point of the circle. The time complexity
is O(N) and the space complexity is O(1).

Another way is to use HasSet to record each node, the first node that exist in hashset would be the start
point. The time complexity is O(N) and space complexity is O(N)
*/

import java.util.*;

public class CircleListStartPoint {
    public Node findStart(Node head) {
        //Pay attention to the first case
        Node slow = head, fast = head;
        do {
            slow = slow.next;
            fast = fast.next.next;
        }while(slow != fast);
        Node temp = head;
        while(slow != temp) {
            slow = slow.next;
            temp = temp.next;
        }
        return slow;
    }

    public Node findStart2(Node head) {
        HashSet<Node> set = new HashSet<Node>();
        Node temp = head;
        while(true) {
            if(set.contains(temp) == false) {
                set.add(temp); temp = temp.next;
            }
            else {
                return temp;
            }
        }
    }

    public static void main(String[] args) {
        Node node1 = new Node(1); Node node2 = new Node(2); Node node3 = new Node(3); 
        Node node4 = new Node(4); Node node5 = new Node(5); Node node6 = new Node(6);
        node1.next = node2; node2.next = node3; node3.next = node4; node4.next = node5;
        node5.next = node6; node6.next = node1;
        CircleListStartPoint clsp = new CircleListStartPoint();
        System.out.println(clsp.findStart(node1).val); 
    }
}