链表中的节点每k个一组翻转 C++

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    ListNode* reverseKGroup(ListNode* head, int k) {
        // write code here
        ListNode* Cur = head;
        int iCount = 0;//元素总个数
        while (Cur) {
            iCount++;
            Cur = Cur->next;          
        }

        double iNum = iCount/k;//翻转组数 
        ListNode* List = head;
        for (int i = 0; i < iNum; i++) {
            head = reverseBetween(List, i*k+1, (i+1)*k);
            List = head;
        }
        return head;
    }

  //翻转指定m-n的节点
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode* Cur = head;
        ListNode* Next = nullptr;
        ListNode* Nodem = head;
        ListNode* Noden = head;
        
        //保存m和n原位置的地址
        for(int i = 0; i < m-1; i++) {                 
            Nodem = Nodem->next;
        }    

        for(int i = 0; i < n; i++) {                 
            Noden = Noden->next;
        }    

        //反转区间链表
        Cur = Nodem;
        ListNode* Pre = Noden; 
        for(int i = 0 ; i <= n-m ; i++){     
            Next = Cur->next;
            Cur->next = Pre;
            Pre = Cur; 
            Cur = Next;
        }
 
        //三表相连接
        if (m == 1) {
            return Pre;
        }

        Cur = head;
        for(int i = 1; i < m-1; i++) {                 
            Cur = Cur->next;
        }      
        Cur->next = Pre;

        return head;
    }
};