Reverse Nodes in k-Group

翻转节点

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

解：

/**
* 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) {
if(head==NULL || k<=1)
return head;

ListNode * myHead=new ListNode(0);//重点：新增一个节点，方便操作
myHead->next=head;

ListNode *priv=myHead, *cur, *tmp, *curNext, *curPriv, *curTail;
int count=0;
while(head){
count++;
if(count==k){
cur=priv->next->next;//记录当前节点
curPriv=priv->next;//记录当前节点的上一个节点
curTail=priv->next;//记录本趟翻转的最后一个节点
curNext=head->next;//记录下一个节点
while(cur!=curNext){
tmp=cur->next;
cur->next=curPriv;
curPriv=cur;
cur=tmp;
}
priv->next=curPriv;
priv=curTail;
priv->next=curNext;
count=0;
head=curNext;
continue;
}
head=head->next;
}
head=myHead->next;

return head;
}
};