D3 链表中的节点每k个一组翻转

https://www.nowcoder.com/practice/b49c3dc907814e9bbfa8437c251b028e?tpId=117&rp=1&ru=%2Fexam%2Fcompany&qru=%2Fexam%2Fcompany&sourceUrl=%2Fexam%2Fcompany&difficulty=&judgeStatus=&tags=&title=&gioEnter=menu

代码

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    ListNode* reverseKGroup(ListNode* head, int k) {
        auto tail = head;
        //判断k节车厢是否满载
        for(int i = 0;i < k;i++){
            if(!tail) return head;
            tail = tail->next;
        }
        //反转该k节车厢
        ListNode* res = reverse(head, tail);
        //继续反转后面的车厢 并且连接(注意此时head为该节车厢的尾部)
        head->next = reverseKGroup(tail, k);
        return res;
    }
    ListNode* reverse(ListNode* cur,ListNode* tail){
        auto pre = tail;
        while(cur != tail){
            auto post = cur->next;
            cur->next = pre;
            pre = cur;
            cur = post;
        }
        return pre;
    }
};