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;
    }
}