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 class Solution {
/**
* @param head: The first node of linked list.
* @param n: An integer.
* @return: Nth to last node of a singly linked list.
*/
ListNode nthToLast(ListNode head, int n) {
// write your code here
ListNode fast = head;
for(int i = 0; i < n; i++)
fast = fast.next;

ListNode slow = head;

while(fast != null){
fast = fast.next;
slow = slow.next;
}

return slow;
}
}