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) {
        // write code here
        if(head == NULL)return NULL;
        ListNode* tail = head;
        for(int i = 0; i < k; i++) {
            if(tail == nullptr) {
                tail = head; break;
            }
            tail = tail -> next;
        }
        ListNode* pre = NULL;
        ListNode* cur = head;
        while(cur != tail) {
            ListNode* tmp = cur -> next;
            cur -> next = pre;
            pre = cur;
            cur = tmp;
        }
        if(tail != head) {
            head -> next = reverseKGroup(tail, k);
        }
        else {
            pre = head;
        }
        return pre;
    }
};