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
        ListNode dummyListNode = new ListNode(-1);
        dummyListNode.next = head;
        ListNode pre = dummyListNode;
        int i = 0;
        ListNode leftNode = pre.next;
        ListNode rightNode = pre.next;
        ListNode tmpNode;
        while(rightNode != null){
            i++;
            if(i == k){
                i =0;
                tmpNode = rightNode.next;
                pre.next = null;
                rightNode.next = null;
                reverseNode(leftNode);
                pre.next = rightNode;
                leftNode.next = tmpNode;
                pre = leftNode;
                rightNode = tmpNode;
                leftNode = tmpNode;
            }else{
                rightNode = rightNode.next;
            }
        }
        return dummyListNode.next;
    }

    public ListNode reverseNode(ListNode head){
        ListNode newListNode = null;
        while(head != null){
            ListNode tmpNode = head.next;
            head.next = newListNode;
            newListNode = head;
            head = tmpNode;
        }
        return newListNode;
    }
}