LintCode Nth to Last Node in List 链表倒数第n个节点

找到单链表倒数第n个节点，保证链表中节点的最少数量为n。

样例

给出链表 3->2->1->5->null和n = 2，返回倒数第二个节点1.

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

The minimum number of nodes in list is n.

Example

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) {
ListNode p = head, q = head;
while(n > 0 && p != null) {
p = p.next;
n--;
}
while(p != null) {
q = q.next;
p = p.next;
}
return q;
}
}