leetcode-Reverse Nodes in k-Group

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

思路：每k个翻转一下，（翻转函数不太熟练）

代码：

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;
}


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;
}

再附一个单链表反转的函数（思路：一个节点一个节点的反转）

ListNode *reverse(ListNode *L)
{
ListNode *cur=L->next;
ListNode *pre=L;
ListNode *head=L;
while(cur!=NULL)
{
ListNode *next=cur->next;

cur->next=pre;
pre=cur;
cur=next;
}
head->next=NULL;
return pre;
}