题解 | #链表中的节点每k个一组翻转#

先进行链表的长度的遍历，然后计算总共的逆序轮数，最后余下的直接接在新的链表的末尾即可。

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    ListNode* reverseKGroup(ListNode* head, int k) {
        // write code here
        ListNode* newhead = (ListNode*)malloc(sizeof(ListNode));
        newhead->next = NULL;
        newhead->val = -1;
        ListNode *curhead = newhead,*pre = NULL,*temp,*curnext;
        pre = head;
        int cnt = 0;
        while(pre){
            cnt++;
            pre = pre->next;
        }
        //判断次数与剩下节点数
        int circle = cnt / k;
        int mod = cnt - circle * k;
        pre = NULL;
        while(circle--){
            for(int i = 1;i <= k;++i){
                temp = curhead->next;
                curnext = head->next;
                curhead->next = head;
                if(pre == NULL)
                    pre = head;
                head->next = temp;
                head = curnext;
            }
            curhead = pre;
            pre = NULL;
        }
        if(mod)
            curhead->next = head;
        return newhead->next;
    }
};