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

/** * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (head == NULL) return NULL;                ListNode* last = new ListNode(0);        last->next = head;        head = last;        while(last->next != NULL)        {            // p1,p2指向交换的两个节点            ListNode* p1 = last->next;            ListNode* p2 = p1;            int count = 1;            while (count < k && p2->next != NULL)            {                p2 = p2->next;                ++count;            }            if (count == k)            {                // p0记录交换子链的第一个节点                ListNode* p0 = p1;                p2 = p1->next;                while(count-- > 1)                {                    ListNode* tmp = p2->next;                    p2->next = p1;                    p1 = p2;                    p2 = tmp;                }                last->next = p1;                p0->next = p2;                last = p0;            }            else            {                break;            }        }        return head->next;    }};