leetcode链表递归-链表反转2

/**
给你单链表的头指针 <code>head</code> 和两个整数 <code>left</code> 和 <code>right</code> ，其中 <code>left <= right</code> 。请你反转从位置 <code>left</code> 到位置 <code>right</code> 的链表节点，返回 <strong>反转后的链表</strong> 。
<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/rev2ex2.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>输入：</strong>head = [1,2,3,4,5], left = 2, right = 4
<strong>输出：</strong>[1,4,3,2,5]
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>head = [5], left = 1, right = 1
<strong>输出：</strong>[5]
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>链表中节点数目为 <code>n</code></li>
	<li><code>1 <= n <= 500</code></li>
	<li><code>-500 <= Node.val <= 500</code></li>
	<li><code>1 <= left <= right <= n</code></li>
</ul>

<p> </p>

<p><strong>进阶：</strong> 你可以使用一趟扫描完成反转吗？</p>
<div><div>Related Topics</div><div><li>链表</li></div></div><br><div><li>👍 1239</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 {
	//后驱节点
	//反转区间内节点
	ListNode successor = null;
    public ListNode reverseBetween(ListNode head, int left, int right) {
		if(left==1){
			return reverseN(head,right);
		}
		head.next=  reverseBetween(head.next,left-1,right-1);
		return head;
    }

	
	//反转前n个节点
	ListNode reverseN(ListNode head,int n){
		if(n==1){
			successor = head.next;
			return head;
		}

		ListNode last = reverseN(head.next,n-1);
		head.next.next = head;
		head.next = successor;
		return last;
	}
}
//leetcode submit region end(Prohibit modification and deletion)