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

解题思路：这题的思路是用stack将每次调换顺序的group先保存下来，一组一组的处理,最后一组不足k值的stack，则不做换序处理.代码写的有点乱，不过勉强AP了.

#include<iostream>
#include<vector>
#include<stack>
using namespace std;

//Definition for singly - linked list.
struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};
ListNode *reverseKGroup(ListNode *head, int k) {
	ListNode*ResultList = new ListNode(0);
	ListNode*Result_tmpnode = ResultList;
	stack<ListNode*>GroupNode;
	ListNode*Tmphead = head;

	int GroupCount = 0; //用来统计每组GROUP的个数.
	bool Split = true;  //最后一组不满k的情况

	while (Tmphead!=NULL){
		GroupCount = 0;
		for (; GroupCount < k; ++GroupCount){
			if (Tmphead == NULL){       //不满k个
				Split =false;
				for (; GroupNode.size() != 1; GroupNode.pop()){}
				Result_tmpnode->next = GroupNode.top();
				break;
			}
			GroupNode.push(Tmphead);
			Tmphead = Tmphead->next;
		}	
		while (!GroupNode.empty()&&Split){
			Result_tmpnode->next = GroupNode.top();
			GroupNode.pop();
			Result_tmpnode = Result_tmpnode->next;
		}
	}
	if (Split)
		Result_tmpnode->next = NULL;
	return ResultList->next;
}

Reverse Nodes in k-Group