25. K 个一组翻转链表

思路很简单，就是头插法翻转cnt次链表就可以了 

class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode*temp=head;
        int len=0;
        while(temp!=nullptr){
            len++;
            temp=temp->next;
        }
        int cnt=len/k;
        ListNode*dummy=new ListNode(-1,head);
        ListNode*fast=head;
        ListNode*slow=dummy;
        for(int i=0;i<cnt;i++){
            for(int j=1;j<k;j++){
                ListNode*temp1=fast->next;
                fast->next=temp1->next;
                temp1->next=slow->next;
                slow->next=temp1;
            }
            slow=fast;
            fast=slow->next;
        }
        return dummy->next;
    }
};

 

class Solution {
public:
    ListNode*myreverse(ListNode*head,ListNode*end){
        ListNode*cur=head;
        ListNode*pre=nullptr;
        while(cur!=end){
            ListNode*nxt=cur->next;
            cur->next=pre;
            pre=cur;
            cur=nxt;
        }
        return pre;
    }
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode*cur=head;
        for(int i=0;i<k;i++){
            if(cur!=nullptr)
                cur=cur->next;
            else
                return head;
        }
        ListNode*newhead=myreverse(head,cur);
        head->next=reverseKGroup(cur,k);
        return newhead;
    }
};

递归思维：k 个一组反转链表