leetcode链表递归-反转链表

/**
给你单链表的头节点 <code>head</code> ，请你反转链表，并返回反转后的链表。
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<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/rev1ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>输入：</strong>head = [1,2,3,4,5]
<strong>输出：</strong>[5,4,3,2,1]
</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/rev1ex2.jpg" style="width: 182px; height: 222px;" />
<pre>
<strong>输入：</strong>head = [1,2]
<strong>输出：</strong>[2,1]
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>head = []
<strong>输出：</strong>[]
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>链表中节点的数目范围是 <code>[0, 5000]</code></li>
	<li><code>-5000 <= Node.val <= 5000</code></li>
</ul>

<p> </p>

<p><strong>进阶：</strong>链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题？</p>
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<div><div>Related Topics</div><div><li>递归</li><li>链表</li></div></div><br><div><li>👍 2453</li><li>👎 0</li></div>
*/

//leetcode submit region begin(Prohibit modification and deletion)
/**
 * 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 {
    //不知道该说什么，大佬牛逼
    //每一次调转前后指针的顺序，从最后俩个开始
    public ListNode reverseList(ListNode head) {
        if(head == null || head.next==null){
            return head;
        }

        ListNode last = reverseList(head.next);
        head.next.next = head;
        head.next = null;
        return last;
    }
}
//leetcode submit region end(Prohibit modification and deletion)