Lintcode-easy-nth to last Node in List Show result

Find the nth to last element of a singly linked list.

The minimum number of nodes in list is n.

Given a List 3->2->1->5->null and n = 2, return node whose value is 1.

/*** Definition for ListNode. * public class ListNode {* int val; * ListNode Next; * ListNode (int val) {* This.val = val; * This.next = null; *     } * } */  Public classSolution {/**     * @paramhead:the first node of linked list. * @paramN:an Integer. * @return: Nth to the last node of a singly linked list. */ListNode Nthtolast (ListNode head,intN) {//Write your code hereListNode fast =Head;  for(inti = 0; I < n; i++) Fast=Fast.next; ListNode Slow=Head;  while(Fast! =NULL) {Fast=Fast.next; Slow=Slow.next; }                returnslow; }}

Lintcode-easy-nth to last Node in List Show result