题解 | #链表中的节点每k个一组翻转#
链表中的节点每k个一组翻转
https://www.nowcoder.com/practice/b49c3dc907814e9bbfa8437c251b028e
import java.util.*; /* * public class ListNode { * int val; * ListNode next = null; * public ListNode(int val) { * this.val = val; * } * } */ public class Solution { /** * 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 */ public ListNode reverseKGroup (ListNode head, int k) { // write code here ListNode dummy = new ListNode(-1); dummy.next = head; ListNode pre = dummy; ListNode end = dummy; while(end != null){ for(int i = 0;i<k && end != null;i++) end = end.next; if(end == null) break; ListNode next = end.next; ListNode start = pre.next; end.next = null; pre.next = reverse(start); start.next = next; pre = start; end = start; } return dummy.next; } public ListNode reverse(ListNode head){ ListNode pre = null; ListNode cur = head; while(cur != null){ ListNode next = cur.next; cur.next = pre; pre = cur; cur = next; } return pre; } }