leetcode連結清單遞歸-反轉連結清單

/**
給你單連結清單的頭節點 <code>head</code> ，請你反轉連結清單，并傳回反轉後的連結清單。
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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>輸入：</strong>head = [1,2,3,4,5]
<strong>輸出：</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>輸入：</strong>head = [1,2]
<strong>輸出：</strong>[2,1]
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>輸入：</strong>head = []
<strong>輸出：</strong>[]
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
    <li>連結清單中節點的數目範圍是 <code>[0, 5000]</code></li>
    <li><code>-5000 <= Node.val <= 5000</code></li>
</ul>

<p> </p>

<p><strong>進階：</strong>連結清單可以選用疊代或遞歸方式完成反轉。你能否用兩種方法解決這道題？</p>
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<div><div>Related Topics</div><div><li>遞歸</li><li>連結清單</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 {
    //不知道該說什麼，大佬牛逼
    //每一次調轉前後指針的順序，從最後倆個開始
    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)      

不會，我可以學；落後，我可以追趕；跌倒，我可以站起來！