题解 | #链表中的节点每k个一组翻转#

/**
 * 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) {
        // write code here
        //1.group
        int kk = k;
        ListNode* p = head;
        ListNode* tail = nullptr;

        while (p) {
            ListNode* q = p;
            while (kk > 1 && q->next) {
                q = q->next;
                kk--;
            }

            if (kk == 1) {
                kk = k;
                ListNode* temp = q->next;
                q->next = nullptr;
                if (tail) {
                    tail->next = reverse(p);
                } else {
                    // the head after reverse
                    head = reverse(p);
                }
                tail = p;
                p = temp;
            } else {
                // keep original type that no satisfy k group
                if (tail) {
                    tail->next = p;
                }
                break;
            }
        }
        return head;
    }

    ListNode* reverse(ListNode* head) {
        ListNode* ptr = nullptr;
        while (head) {
            ListNode* next = head->next;
            head->next = ptr;
            ptr = head;
            head = next;
        }
        return ptr;
    }
};