要实现按照 k 个一组反转链表的功能，先将链表按每 k 个节点一组进行划分。对每一组节点进行反转操作。将反转后的组重新连接起来。链表 1->2->3->4->5->6->7，然后需要按照k个一组反转比如k=2，那么需要得到2->1->4->3->6->5->7k=3，得到3->2->1->6->5->4->7```// 定义链表节点类class ListNode { int val; ListNode next; ListNode(int val) { this.val = val; }}public class ReverseNodesInKGroup { public ListNode reverseKGroup(ListNode head, int k) { if (head == null || k == 1) { return head; } // 创建一个虚拟头节点，方便处理 ListNode dummy = new ListNode(0); dummy.next = head; ListNode prev = dummy; ListNode end = dummy; while (end.next != null) { // 找到当前组的最后一个节点 for (int i = 0; i < k &amp;&amp; end != null; i++) { end = end.next; } if (end == null) { break; } ListNode start = prev.next; ListNode nextGroup = end.next; end.next = null; // 反转当前组的节点 prev.next = reverseList(start); start.next = nextGroup; prev = start; end = start; } return dummy.next; } // 反转链表的方法 private ListNode reverseList(ListNode head) { ListNode prev = null; ListNode current = head; while (current != null) { ListNode nextNode = current.next; current.next = prev; prev = current; current = nextNode; } return prev; } public static void main(String[] args) { // 创建链表 1->2->3->4->5->6->7 ListNode head = new ListNode(1); ListNode node2 = new ListNode(2); ListNode node3 = new ListNode(3); ListNode node4 = new ListNode(4); ListNode node5 = new ListNode(5); ListNode node6 = new ListNode(6); ListNode node7 = new ListNode(7); head.next = node2; node2.next = node3; node3.next = node4; node4.next = node5; node5.next = node6; node6.next = node7; ReverseNodesInKGroup solution = new ReverseNodesInKGroup(); int k = 2; ListNode newHead = solution.reverseKGroup(head, k); // 打印反转后的链表 while (newHead != null) { System.out.print(newHead.val); if (newHead.next != null) { System.out.print(&quot;->&quot;); } newHead = newHead.next; } }}```
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