Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked 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
  題意：給定一個單鏈表和一個數值，將單鏈表中以該數值個節點為一組進行翻轉（不可以只交換節點值）

/**
 * 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 == nullptr || k == 1) return head;
        int num = 0;
        ListNode *preheader = new ListNode(-1);
        preheader->next
 
 = head;
        ListNode *cur = preheader, *nex, *pre = preheader;
        while (cur = cur->next) num++; //連結串列長度
        while (num >= k) {
            cur = pre->next;
            nex = cur->next;
            for (int i = 1; i < k; ++i) {
                cur->next = nex->next;
                nex->next = pre->next;
                pre->next = nex;
                nex = cur->next;
            }//翻轉即每次從cur指向的節點後面取一個節點放到pre後面，pre指向第一個需要翻轉節點之前的節點
            pre = cur;
            num -= k;
        }
        return preheader->next;
    }
};