题目来源：

(题目来源 (opens new window) " 206.【简单】反转链表")

题目

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

进阶: 你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

## 分析

这个关键就是查找


## 题解

方法一
```java

// @lc code=start
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if (n <= 0 || head == null) {
            return head;
        }


        ListNode cur = head;
        int size = 0;
        while (cur != null) {
            size++;
            cur = cur.next;
        }
        int num = size - n;

        ListNode pre = new ListNode(-1);
        pre.next = head;
        cur = pre;

        while(num > 0) {
            cur = cur.next;
            -- num;
        }

        cur.next = cur.next.next;
        return pre.next;
    }
}

输入
[1,2,3,4,5] 
1
输出
[1,2,3,4]

递归解法

class Solution2 {
    public ListNode reverseList(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode rev = reverseList(head.next);
        head.next.next = head;
        head.next = null;
        return rev;
    }
}