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 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
        if(head == null || k == 1 )
        {
            return head;
        }
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode pre = dummy;
        ListNode end = dummy;
        while(true)
        {
        for(int i = 0 ; i < k && end != null ; i ++)
        {
            end = end.next;
        }
        if(end == null)
        {
         break;   
        }
        ListNode start = pre.next;
        ListNode next = end.next;
        end.next = null;
        pre.next = reverse(start);
        start.next = next;
        pre = start;
        end = pre;
        }
        System.out.print(dummy.next);
        return dummy.next;
    }
    private ListNode reverse(ListNode head)
    {
        ListNode prev = null;
        ListNode curr = head;
        while(curr != null)
        {
            ListNode nextTemp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextTemp;
        }
        return prev;
    }
}