25

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        ListNode end = dummy;

        while(end.next != 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;//将k段分出去以后需要在后面添上一个null已断开连接
            pre.next = reverse(start);
            start.next = next;
            pre = start;
            end = start;
        }
        return dummy.next;
    }

    public static ListNode reverse(ListNode start){
        ListNode pre = null;
        ListNode cur = start;
        while(cur  != null){
            ListNode next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
}

来源：https://www.cnblogs.com/zhaijiayu/p/11521275.html