给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。
示例 1：
输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]
示例 2：
输入：head = [5], left = 1, right = 1
输出：[5]
//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 {
public ListNode reverseBetween(ListNode head, int left, int right) {
if (head == null || head.next == null || left == right) {
return head;
}
//虚拟节点为了避免头节点为空
ListNode dummy = new ListNode(-1);
dummy.next = head;
//分别记录要反转的前节点和要反转的第一个节点
ListNode currDummy = dummy;
ListNode prevDummy = null;
ListNode prev = null;
ListNode curr = null;
for (int i = 0; i < left; i++) {
prevDummy = currDummy;
currDummy = currDummy.next;
}
curr = currDummy;
//反转范围内的节点
for (int i = left; i <= right && curr !=null; i++){
ListNode next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
//拼接节点，反转前的结点 拼反转后的头结点
prevDummy.next = prev;
//反转后的最后一个节点拼接right后的那个节点
currDummy.next = curr;
return dummy.next;
}
}
//leetcode submit region end(Prohibit modification and deletion)