有一说一，这题有点绕。。要找四个点才能解。

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        auto dummy = new ListNode(0, head);

        //pre是反转开始节点的前一个节点，st为反转开始节点
        //ed是反转最后一个节点，after是反转最后节点的后一节点
        auto pre = dummy, ed = dummy;

        for (int i = 0; i < m - 1; i ++ ) pre = pre->next;
        for (int i = 0; i < n; i ++ ) ed = ed->next;

        auto st = pre->next, after = ed->next;

        auto p1 = st, p2 = st->next;    //这一段是反转链表模板
        while (p2 != after)
        {
            auto p3 = p2->next;
            p2->next = p1;
            p1 = p2, p2 = p3;
        }

        pre->next = ed;
        st->next = after;

        return dummy->next;
    }
};