题解 | #牛群的重新分组#

/**
 * 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 dummy = new ListNode(-1);//创建临时头结点
        dummy->next = head;
        ListNode* pre = dummy;
        ListNode* cur = head;
        while (cur != nullptr) {
            //保存每一组的链表头部,类似带头结点的局部链表
            ListNode* ehead = pre;
            //组内成员不足k个的处理
            ListNode* flag = cur;
            for (int i = 1; i < k; i++) {
                flag = flag->next;
                if (flag == nullptr) {
                    break;
                }
            }
            //组内有k个元素时才反转
            if (flag) {
                //每k个一组链表反转
                for (int i = 1; i <= k; i++) {
                    ListNode* temp = cur->next;
                    cur->next = pre;
                    pre = cur;
                    cur = temp;
                }
                ehead->next->next = cur;
                ehead->next = pre;
                for (int i = 1; i < k; i++) {
                    pre = pre->next;//更新pre为新的前驱
                }
            } else {
                break;
            }
        }
        return dummy->next;
    }
};