leetcode Reverse Nodes in k-Group

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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

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *reverseKGroup(ListNode *head, int k) {        int len=0;        ListNode *lencount=head;        while(lencount!=NULL) {            len++;            lencount=lencount->next;        }        if(len<=1||k==1) return head;        ListNode *result=head,*pre,*start=head,*next,*headTemp,*tailTemp;        bool flag=false;        while((len=len-k)>=0){            if(!flag){                next=sub(start,k,result,pre);                flag=true;            }            else{                start=next;                next=sub(start,k,headTemp,tailTemp);                pre->next=headTemp;                pre=tailTemp;            }        }        pre->next=next;        return result;    }    ListNode* sub(ListNode *start,int k,ListNode *&head,ListNode *&tail){     //逆转start开头的K个node,并且保存逆转后的开头和结尾，并返回下一个node，以供连接        ListNode *sting=new ListNode(INT_MIN),*pre,*cur,*next;        int len=0;        tail=start;        sting->next=start;        pre=sting;        cur=start;        while(len<k){            next=cur->next;            cur->next=pre;            pre=cur;            cur=next;            len++;        }        head=pre;        return next;    }};

开始的时候感觉遍历两边处理就不是很好，果然有巧妙的方法。如下方法，将pre加入到逆转，实现了连接功能

public ListNode reverseKGroup(ListNode head, int k) {      if(head == null)      {          return null;      }      ListNode dummy = new ListNode(0);      dummy.next = head;      int count = 0;      ListNode pre = dummy;      ListNode cur = head;      while(cur != null)      {          count ++;          ListNode next = cur.next;          if(count == k)          {              pre = reverse(pre, next);              count = 0;             }          cur = next;      }      return dummy.next;  }  private ListNode reverse(ListNode pre, ListNode end)  {      if(pre==null || pre.next==null)          return pre;      ListNode head = pre.next;      ListNode cur = pre.next.next;      while(cur!=end)      {          ListNode next = cur.next;          cur.next = pre.next;          pre.next = cur;          cur = next;      }      head.next = end;      return head;  }

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