K组翻转链表-LintCode

给你一个链表以及一个k,将这个链表从头指针开始每k个翻转一下。
链表元素个数不是k的倍数，最后剩余的不用翻转。

样例：
给出链表 1->2->3->4->5
k = 2, 返回 2->1->4->3->5
k = 3, 返回 3->2->1->4->5

#ifndef C450_H
#define C450_H
#include<iostream>
#include<stack>
using namespace std;
class ListNode{
public:
    int val;
    ListNode *next;
    ListNode(int val)
    {
        this->val = val;
        this->next = NULL;
    }
};
class Solution {
public:
    /* * @param head: a ListNode * @param k: An integer * @return: a ListNode */
    ListNode * reverseKGroup(ListNode * head, int k) {
        // write your code here
        if (head == NULL || k == 1)
            return head;
        ListNode *p = head;
        int num = 0;
        while (p != NULL)
        {
            num++;
            p = p->next;
        }
        if (k > num)
            return head;
        stack<int> sk;
        ListNode *q = head;
        ListNode *node = new ListNode(-1);
        ListNode *t = node;
        int n = 1;
        while (q != NULL)
        {
            if (n%k != 0)
            {
                sk.push(q->val);
            }
            else
            {
                t->next = new ListNode(q->val);
                t = t->next;
                while (!sk.empty())
                {
                    t->next = new ListNode(sk.top());
                    sk.pop();
                    t = t->next;
                }
            }
            ++n;
            q = q->next;
        }
        ListNode *l = t->next;
        while (!sk.empty())
        {

            t->next = new ListNode(sk.top());
            t->next->next = l;
            l = t->next;
            sk.pop();
        }
        return node->next;
    }
};
#endif