这个就是用栈，k个一组判断就好了

/*** 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 hereListNode* tmp = head;int cnt = 0;while (tmp) {cnt++;tmp = tmp->next;}int mo = cnt / k;if (mo == 0) {return head;}//stack<ListNode*>s[mo + 5];stack<int>s[mo+5];int res = 1;for (int i = 1; i <= 2000 && res <= mo; i++) {if (i % k != 0) {s[res].push(head->val);head = head->next;} else {s[res].push(head->val);head = head->next;res++;}}ListNode* Head;Head = new ListNode(s[1].top());s[1].pop();tmp = Head;for (int i = 1; i <= mo; i++) {while (!s[i].empty()) {tmp->next = new ListNode(s[i].top());cout << Head->val << endl;//cout<<tmp->val<<endl;tmp = tmp->next;s[i].pop();}}tmp->next = head;return Head;}
};

栈中保存结点

/*** 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 hereListNode* tmp = head;int cnt = 0;while (tmp) {cnt++;tmp = tmp->next;}int mo = cnt / k;if (mo == 0) {return head;}stack<ListNode*>s[mo + 5];//stack<int>s[mo+5];int res = 1;for (int i = 1; i <= 2000 && res <= mo; i++) {if (i % k != 0) {s[res].push(head);head = head->next;} else {s[res].push(head);head = head->next;res++;}}ListNode* Head;Head =s[1].top();s[1].pop();tmp = Head;for (int i = 1; i <= mo; i++) {while (!s[i].empty()) {tmp->next =  s[i].top();//cout << Head->val << endl;//cout<<tmp->val<<endl;tmp = tmp->next;s[i].pop();}}tmp->next = head;return Head;}
};

 递归求解

递归就是栈

/*** 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 hereListNode *tail=head;for(int i=0;i<k;i++){if(tail==nullptr){return head;}tail=tail->next;}ListNode *pre=nullptr;ListNode *cur=head;ListNode *tmp=nullptr;while(cur!=tail){tmp=cur->next;cur->next=pre;pre=cur;cur=tmp;}head->next=reverseKGroup(tail, k);return pre;}
};