2.2---找链表倒数第K个结点

答案，注意，一种是递归，另一种是迭代，那么巧妙利用双指针：

迭代：

    public static LinkedListNode nthToLast(LinkedListNode head, int n) {
        LinkedListNode p1 = head;
        LinkedListNode p2 = head;
        
        if (n <= 0) return null;
        
        // Move p2 n nodes into the list.  Keep n1 in the same position.
        for (int i = 0; i < n - 1; i++) { 
            if (p2 == null) {
                return null; // Error: list is too small.
            }
            p2 = p2.next;
        }
        if (p2 == null) { // Another error check.
            return null;
        }
        
        // Move them at the same pace.  When p2 hits the end, 
        // p1 will be at the right element.
        while (p2.next != null) {
            p1 = p1.next;
            p2 = p2.next;
          }
          return p1;
    }

递归：

	public static int nthToLastR1(LinkedListNode head, int n) {
		if (n == 0 || head == null) {
			return 0;
		}
		int k = nthToLastR1(head.next, n) + 1;
		if (k == n) {
			System.out.println(n + "th to last node is " + head.data);
		}
		return k;
	}