来源：力扣（LeetCode） 链接：https://leetcode.cn/problems/reverse-linked-list-ii 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

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 。

解答

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} left
 * @param {number} right
 * @return {ListNode}
*/
const reverseChain = (head) => {
    let prev = null;
    while (head) {
        const next = head.next;
        head.next = prev;
        prev = head;
        head = next;
    }
}
var reverseBetween = function(head, left, right) {
    const firstNode = new ListNode(null, head);
    let curNode = firstNode,
        leftStartNode = null,
        rightStartNode = null,
        leftLastNode = null,
        rightLastNode = null,
        idx = 0;
    while (curNode) {
        if ((idx + 1) === left) {
            leftStartNode = curNode?.next;
            leftLastNode = curNode;
        }
        if (idx === right) {
            rightStartNode = curNode;
            rightLastNode = curNode.next;
            if (leftStartNode !== null) {
                rightStartNode.next = null;
                reverseChain(leftStartNode);
                leftLastNode.next = rightStartNode;
                leftStartNode.next = rightLastNode;
            }
        }
        ++idx;
        curNode = curNode.next;
    }
    return firstNode.next;
};