题解 | #链表中的节点每k个一组翻转(又臭又长版)#

/**
 * 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) {
        if (head == nullptr) return head;

        bool reverseflag = true;

        ListNode* res = nullptr;
        ListNode* p1 = head;

        vector<ListNode*> listvector;

        while (true) {

            cout << "p1:" << p1->val << endl;
            ListNode* cur = p1;
            //判断是否反转
            ListNode* cur1 = p1;
            int j = 0;
            for (int i = 0; i < k - 1; i++) {
                if (cur1 -> next != nullptr) cur1 = cur1->next;
                else break;
                j++;
            }
            if (j == k - 1) reverseflag = true;
            else reverseflag = false;

            res = nullptr;
            if (reverseflag) {
                for (int i = 0; i < k; i++) {
                    ListNode* newnode = new ListNode(cur->val);
                    newnode->next = res;
                    res = newnode;
                    cur = cur->next;
                }
            } else {
                res = p1;
            }
            listvector.push_back(res);

            for (int i = 0; i < k; i++) {
                p1 = p1->next;
                if (p1 == nullptr) break;
            }
            if (p1 == nullptr) break;
        }

        for (int i = 0; i < listvector.size() - 1; i++) {
            ListNode* temp = listvector[i];
            ListNode* temp1 = listvector[i + 1];
            while (temp->next != nullptr) {
                temp = temp->next;
            }
            temp->next = temp1;
        }

        res = listvector[0];
        return res;
    }
};