move 1 from A to C
move 2 from A to B
move 1 from C to B
move 3 from A to C
move 1 from B to A
move 2 from B to C
move 1 from A to C

move 1 from A to B
move 2 from A to C
move 1 from B to C
move 3 from A to B
move 1 from C to A
move 2 from C to B
move 1 from A to B
move 4 from A to C
move 1 from B to C
move 2 from B to A
move 1 from C to A
move 3 from B to C
move 1 from A to B
move 2 from A to C
move 1 from B to C

    /**
     * @param topNs 盘子总数
     * @param from     盘子原来所在的柱子
     * @param inter     辅助用的柱子
     * @param to     盘子要移动到的柱子
     */
    public static void move(int topNs, char from, char inter, char to) {
        if (topNs == 1) {
            System.out.printf("move %s from %s to %s\n", topNs, from, to);
        } else {
            move(topNs - 1, from, to, inter);
            System.out.printf("move %s from %s to %s\n", topNs, from, to);
            move(topNs - 1, inter, from, to);
        }
    }

------解决方案--------------------
以前一直不会解这个题目，后来老师给了一句口诀，然后就很简单的会了：
老和尚先叫小和尚把上面的N-1个盘子搬到C柱，然后老和尚再把最后的盘子搬到C柱

------解决方案--------------------
看过《猩球崛起》吗？里面人类用大猩猩做实验就有这道题目。

看来，程序员还是没能逃脱猿猴的命运啊......

------解决方案--------------------
汉诺塔问题啊，递归解决

------解决方案--------------------
探讨


看过《猩球崛起》吗？里面人类用大猩猩做实验就有这道题目。

看来，程序员还是没能逃脱猿猴的命运啊......


------解决方案--------------------
B is temp ，

move num from A to B（所有） go  

move num from B to C

success

------解决方案--------------------
import java.awt.*;  
public class TowerPoint //公共类TowerPoint
{  
int x,y; //定义2个int类型的变量
boolean 有盘子; //定义一个boolean类型的变量
Disk 盘子=null; //初始化一个对象"盘子"并赋值为空
HannoiTower con=null; //初始化一个HannoiTower类的对象"con"并赋值为空

public TowerPoint(int x,int y,boolean boo) //构造函数，有3个参数，x，y，boo
{  
this.x=x; //将参数赋给当前x
this.y=y; //将参数赋给当前y

有盘子=boo; //将boo赋给"有盘子"
}  
public boolean 是否有盘子() //定义一个返回boolean类型的方法"是否有盘子"
{  
return 有盘子; //返回boolean类型的"有盘子"
}  
public void set有盘子(boolean boo) //set方法，并且参数为boolean
{  
有盘子=boo; //将boo赋给有盘子
}  

public int getX() //取得x方法
{