class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        auto dummy = new ListNode(-1, head);
        for (auto p = dummy;;) { //注意这里必须要有两个 ; ; 
            auto q = p;
            for (int i = 0; i < k && q; i ++) q = q->next; //是否足k个节点
            if (q == NULL) break; //说明不足k个节点
            auto a = p->next, b = a->next;
            for (int i = 0; i < k - 1; i ++) {
                auto c = b->next;
                b->next = a; //1 和 2
                // c->next = b; 不要加这句，否则错误
                a = b, b = c; //为2做准备
            }

            auto c = p->next;
            p->next = a, c->next = b; // 分别是3 和 4
            p = c;  //5
        }
        return dummy->next;
    }
};