CTCI 2.2

/*Use two pointers, forward one K nodes, then the two pointers form a "window" contains K nodes.
Then move two pointer one step by one step until the first one reach the end, return the second node.
Pay attention to the situation when k is 0 or k is the length of the linked list, there migth be some 
tricks. 

*/
public class KToLastNode {
    public Node kToLastNode(Node head, int k) {
        Node sent = new Node(0); sent.next = head;
        Node first = sent;
        while(k > 0) {
            first = first.next;
            k--;
        }
        Node second = sent;
        while(first != null) {
            second = second.next;
            first = first.next;
        }
        return second;
    }

    public void print(Node head) {
        Node temp = head;
        while(temp != null) {
            System.out.print(temp.val + " ");
            temp = temp.next;
        }
        System.out.println("");
    }

    public static void main(String[] args) {
        Node node1 = new Node(1); Node node11 = new Node(1); Node node111 = new Node(1);
        Node node2 = new Node(2); Node node22 = new Node(2); Node node3 = new Node(3);
        node1.next = node11; node11.next = node2; node2.next = node111; node111.next = node3; node3.next = node22;
        Node head = node1;
        KToLastNode kt = new KToLastNode();
        kt.print(kt.kToLastNode(head, 2));
    }
}