题目描述

方法一）模拟

class Solution{
public:
	pair<ListNode*, ListNode*> myReverse(ListNode* head, ListNode* tail){
		ListNode* prev = tail->next;
		ListNode* p = head;
		while(prev!=tail){
			ListNode* nex = p->next;
			p->next = prev;
			prev = p;
			p = nex; 
		}
		return {tail, head};
	}
	
	ListNode* reverseKGroup(ListNode* head, int k){
		ListNode* hair = new ListNode(0);
		hair->next = head;
		ListNode* pre = hair;
		while(head){
			ListNode* tail = pre;
			for(int i=0;i<k;i++){
				tail = tail->next;
				if(!tail){
					return hair->next;     //不够k个说明到链表末尾了 
				}
			}
			ListNode* nex = tail->next;
			
			// 翻转长度为k的子链表      这里是 C++17 的写法，也可以写成！！！
            // pair<ListNode*, ListNode*> result = myReverse(head, tail);
            // head = result.first;
            // tail = result.second;
            tie(head, tail) = myReverse(head, tail);

			pre->next = head;  //前后分别连接到原串 
			tail->next = nex;
			pre = tail;       //修改新的子串pre和子串头 
			head = tail->next;
		}
		return hair->next;
	}
};