链表反转及查找倒数第K个节点数据

    /**
     * 实现链表的反转
     * @param head 链表的头节点
     */
    public static void reverseList(HeroNode head){
//      如果链表为空或者只有一个节点,则直接返回即可,无需进行反转
        if(head.next == null || head.next.next == null){
            return;
        }
//      实现链表的反转
        HeroNode temp = head.next;
        HeroNode tm = null;  //周转
        HeroNode heroNode1 = new HeroNode();
        while(temp != null){
            tm = temp.next;
            temp.next = heroNode1.next;
            heroNode1.next = temp;
            temp = tm;
        }
        head.next = heroNode1.next;
    }
    /**
     * 查找倒数第K个节点的下标,并且展示数据
     * @param head 链表的头节点
     * @param index 倒数的第K个节点下标
     * @return 返回第K个节点数据
     */
    public static HeroNode getHeroNode(HeroNode head,int index){
        if (head.next == null){
            System.out.println("链表为空");
            return null;
        }
        if(index > getNum(head)||index <= 0){
            System.out.println("您查询的数据不存在");
            return null;
        }
//      获得有效的节点个数
        int size = getNum(head);
        int no = size - index + 1;
        HeroNode temp = head.next;
        int length = 0;
        while(temp != null){
            length++;
            if (length == no){
                break;
            }
            temp = temp.next;
        }
        return temp;
    }
    /**
     * 静态方法
     * @return 获取节点的个数
     */
    public static int getNum(HeroNode head){
        if (head.next == null){
//          表示没有节点
            return 0;
        }
        HeroNode temp = head.next;
        int length = 0;
        while (temp != null){
            length++;
            temp = temp.next;
        }
        return length;
    }
}