题解 | #链表中的节点每k个一组翻转#

/**
 * 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) {
        if(k==1) return head;
        
        ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        ListNode* pre = dummy;
        ListNode* cur = dummy->next;
        int coun = 0;
        while(cur){
            coun ++;
            cur = cur->next;
        }
        cur = dummy->next;
        int cou = coun/k;
        while( cou-- ){
            int n = k-1;
            while(n--){
                ListNode* tmp = cur->next;
                cur->next = tmp->next;
                tmp->next = pre->next;
                pre->next = tmp;
            }
            pre = cur;
            cur = cur->next;
        }
        return dummy->next;
    }

};