Reverse Nodes in k-Group

描述
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
You may not alter the values in the nodes, only nodes itself may be changed.
Only constant memory is allowed.
For example, Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5
中文：
给定一个链表和一个k值，要求把这个链表每段长度为k的节点反转，不够k值长度的片断不反转。

分析：
根据循环，每遍历k个元素，进行一次翻转。

class Solution {public:   ListNode *reverseKGroup(ListNode *head, int k) {     if (head == nullptr || head->next == nullptr || k < 2)      return head;      ListNode dummy(-1);      dummy.next = head;     for(ListNode *prev = &dummy, *end = head; end; end = prev->next) {            for (int i = 1; i < k && end; i++)                 end = end->next;            if (end == nullptr) break; // 不足k 个            prev = reverse(prev, prev->next, end);//reverse翻转函数，将prev到end的链表翻转                                                 //返回翻转后的倒数第一个元素      }      return dummy.next;   }// prev 是first 前一个元素, [begin, end] 闭区间，保证三者都不为null// 返回反转后的倒数第1 个元素   ListNode* reverse(ListNode *prev, ListNode *begin, ListNode *end) {   ListNode *end_next = end->next;   ListNode *p = begin;   ListNode *cur = p->next;   ListNode *next = cur->next;   while(cur != end_next){       cur->next = p;       p = cur;       cur = next;       next = next ? next->next : nullptr;    }   begin->next = end_next;   prev->next = end;   return begin;}};