题解 | #链表中的节点每k个一组翻转#

//第一步：先看链表长度小于3的，分k=1，k=2直接返回
//第二步：链表长度大于3，做部分反转，每次翻转之后看剩余节点数是否大于等于k，大于等于k继续反转，否则退出循环，结束反转
/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 *  ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
#include <cstddef>
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
        if (head == nullptr || head->next == nullptr) {
            return head;
        }
        if (head->next->next == nullptr) {
            if (k == 2) {
                auto* dummy = new ListNode(0);
                dummy->next = head;
                ListNode* pre = head;
                ListNode* cur = pre->next;
                pre->next = cur->next;
                cur->next = dummy->next;
                dummy->next = cur;
                return dummy->next;
            } else {
                return head;
            }
        }
	  
	  
        auto* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* pre = head;
        ListNode* cur = pre->next;
        ListNode* L = dummy;
        int flag=0;
        while (cur) {
            for (int i = 0; i < k - 1; i++) {
                pre->next = cur->next;
                cur->next = L->next;
                L->next = cur;
                cur = pre->next;
            }
            for(int i=0;i<k-1;i++)
            {
                if(cur==nullptr)
                {
                    flag=1;
                    break;
                }
                cur=cur->next;
            }
            if(flag==1)
            break;
            L = pre;
            pre = L->next;
            cur = pre->next;
        }
        return dummy->next;
    }
};