[lint]. Nth to Last Node in List.java

E
tags: Linked List, Lint

#### Linked List
- 先找到nth node
- 然后head开始跑
- node 到底，而head ~ node刚好是 n 距离。所以head就是要找的last nth

```
/*
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.

Tags Expand 
Cracking The Coding Interview Linked List

Thinking process:
1. Find nth node in normal order.
2. Have a head at index0.
3. Move both head and nth node. WHen nth node hit null/end, then the moving head is the nth to last node in list.
*/

/**
 * 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) {
        if (head == null || n < 0) {
            return null;
        }
        int count = 0;
        ListNode node = head;
        while (node != null && count < n) {
            node = node.next;
            count++;
        }
        while (node != null) {
            node = node.next;
            head = head.next;
        }
        return head;
    }
}




```