反转链表前N个节点

反转链表前N个节点，并返回反转后的链表

ListNode结构如下

    public class ListNode {
        int val;
        ListNode next;
        ListNode(int x) { val = x; }
    }

class Solution {
    //后继节点
    ListNode successor = null;
    public ListNode reverseN(ListNode head, int n){
        if (head == null || head.next == null) return head;
        if (n == 1){
            successor = head.next;
            return head;
        }
        ListNode last = reverseN(head.next, n-1);
        head.next.next = head;
        head.next = successor;
        return last;
    }
}

标签：head,ListNode,反转,next,链表,null,节点,successor
来源： https://blog.csdn.net/qq_40660894/article/details/113476707