题解 | #链表中的节点每k个一组翻转#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    public ListNode reverseKGroup (ListNode head, int k) {
        // write code here
        if(head == null || head.next == null) return head;
        
        ListNode dummy = new ListNode(0);
        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;
            end.next = null;
            
            ListNode start = pre.next;
            pre.next = reverse(start);
            start.next = next;
            
            pre = start;
            end = start;
        }
        return dummy.next;
    }
    
    public ListNode reverse(ListNode head){
        ListNode dummy = new ListNode(0);
        ListNode cur = head;
        while(cur != null){
            ListNode next = cur.next;
            cur.next = dummy.next; //将当前访问的节点 挂到 dummy 的 next 下（头插法）
            dummy.next = cur;
            cur = next;
        }
        return dummy.next;
    }
    
}