BM3 k个一组反转链表

 链表中的节点每k个一组翻转_牛客题霸_牛客网 (nowcoder.com)

/**
 * 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
        //找到每次翻转的尾部
        ListNode* tail = head;
        //遍历k次到尾部
        for(int i = 0; i < k; i++){
            if(tail == nullptr) return head;
            tail = tail->next;
        }   
        //双指针
        ListNode* pre = nullptr;
        ListNode* cur = head;
        //遍历
        while(cur != tail){
            ListNode* temp = cur ->next;
            cur->next = pre;
            pre = cur;
            cur = temp;
        }
        //当前尾指向下一段要翻转的链表
        head->next = reverseKGroup(tail,k);
        return pre;
    }
};