Problem

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.

Note

依然是一道找倒数第n个结点的链表题，用双指针做。fast先走n，然后fast和slow一起走，直到fast为null，slow的位置就是倒数第n个位置。

Solution

public class Solution {
    ListNode nthToLast(ListNode head, int n) {
        ListNode fast = head, slow = head;
        int i = n;
        while (i-- > 0) fast = fast.next;
        while (fast != null) {
            fast = fast.next;
            slow = slow.next;
        }
        return slow;
    }
}