题解 | #链表中的节点每k个一组翻转#

==思路就是==

    一部分一部分翻转，返回头尾指针
最后再一部分一部分拼接起来

/**
 * struct ListNode {
 *    int val;
 *    struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */

    pair<ListNode*, ListNode*> reverselist(ListNode* first, ListNode* second) {
       ListNode* res_1=first;
        ListNode* res_2 = second;

        ListNode* left = nullptr;
        while(first!=second) {
            cout<<first->val<<" ";
            ListNode* tmp = first->next;
            first->next = left;
            left=first;
            first=tmp;

        }
        cout<<second->val<<" ";
        second->next=left;
        cout<<endl;
        return {res_2,res_1};


    }


    ListNode* reverseKGroup(ListNode* head, int k) {
        // write code here
        // 先找到要翻转的区间
        ListNode* first = head;
        ListNode* second=head;
        ListNode* last = head;
        bool flag = true;

        vector<pair<ListNode*, ListNode*>> ve;

        while(last) {
            for(int i=0;i<k-1;i++) {
                if(!second) {
                    break;
                } else {
                    second=second->next;
                }
            }

            if(!second) {
                last=nullptr;
                ve.push_back({first,nullptr});
            } else {
                last=second->next;
                if(flag) {
                    head=second;
                    flag =false;
                }
//                 cout<<"first: "<<first->val<<" second: "<<second->val<<endl;
                auto val = reverselist(first, second);
                ve.push_back(val);
                first=last;
                second=last;

            } 
        }
        // 开始进行拼接
        for(int i=1;i<ve.size();i++) {
            ve[i-1].second->next=ve[i].first;
        }


        return head;


    }
};