题解 | #递归-链表中的节点每k个一组翻转#

/**
 * 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 *tail = head;
        for(int i = 0;i<k;i++){
            if(tail == nullptr) return head; //不足K个的递归那一组不需要翻转，直接返回头结点
            tail = tail->next;
        }
        ListNode *pre = nullptr;
        ListNode *cur = head;
        while (cur != tail) {
            ListNode *temp = cur->next;
            cur->next = pre;
            pre = cur;
            cur = temp;
        }
        head->next = reverseKGroup(tail, k);
        return pre;
    }
};