一个JAVA编写的迷宫算法。。自动找迷宫出口

import java.awt.Color;
import java.awt.GridLayout;
import java.util.Stack;

import javax.swing.JButton;
import javax.swing.JFrame;
import javax.swing.JPanel;

public class FindPath {

private static final int wid = 10;
private static final int hei = 10;
JFrame frame; //窗体,是整个迷宫的容器
JPanel panel;
JButton button[];
private Stack stack = new Stack();
public static void main(String[] args) {

byte[] map ={
1,1,1,1,1,1,1,1,1,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,0,0,1,1,0,0,1,
1,0,1,1,1,0,0,0,0,1,
1,0,0,0,1,0,0,0,0,1,
1,0,1,0,0,0,1,0,0,1,
1,0,1,1,1,0,1,1,0,1,
1,1,0,0,0,0,0,0,0,1,
1,1,1,1,1,1,1,1,1,1,
};

int i=0;

FindPath ai = new FindPath();
ai.find(map, 11, 88);

}

FindPath()
{

int i;
frame = new JFrame("孙尧的神奇迷宫");
frame.setBounds(300,240,500,500); //调整迷宫出现的位置及大小
frame.setResizable(false);//窗体不可拉伸

panel = new JPanel();
frame.getContentPane().add(panel); //将面板添加到窗体中
panel.setLayout(new GridLayout(10,10)); //panel用网格布局,10行10列

button = new JButton[100];
for ( i = 0; i < button.length; i++)
{
button[i]=new JButton(Integer.toString(i)); //创建按键的名字，Integer型的名字为i的字符串
if((i>=0&&i<=9)||(i>=90&&i<=99)||i%10==0||i%10==9||i==13||i==17||i==23||i==27||i==35||i==36||i==42||i==43||i==44||i==54||i==62||i==66||i==72||i==73||i==74||i==76||i==77||i==81)
{
button[i].setBackground(Color.red);

}
else
{
button[i].setBackground(Color.white);

}
button[i].setSize(10,10);
panel.add(button[i]);
}

frame.setDefaultCloseOperation(frame.EXIT_ON_CLOSE); //声明点“X”图标后结束窗体所在的应用程序
frame.setVisible(true); //表明以上创建的所有窗体、按键等组件都是可见的
}

public void find(byte[] map, int origin, int target) {
int[] step = new int[2];
step[1] = origin;
stack.addElement(step);
if(findPath(map,origin,target)){
System.out.println("succ");
for (int i = 0; i < stack.size(); i++) {
int[] temp = (int[])stack.elementAt(i);
System.out.println(i+" /t"+(char)temp[0]+" "+temp[1]);
button[temp[1]].setBackground(Color.blue);
}
}else{
System.out.println("fail");
}
}

public boolean findPath(byte[] map, int origin, int target) {
if (canMoveTo(map, origin, target, 'l')) {
return true;
}
if (canMoveTo(map, origin, target, 'r')) {
return true;
}
if (canMoveTo(map, origin, target, 'u')) {
return true;
}
if (canMoveTo(map, origin, target, 'd')) {
return true;
}
stack.pop();// 如果四个方向都试过，全部不行，那么把当前步骤弹出
return false;
}

private boolean canMoveTo(byte[] map, int origin, int target, char direct) {
int next = 0;
switch (direct) {
case 'l':
next = origin - 1;
break;
case 'r':
next = origin + 1;
break;
case 'u':
next = origin - wid;
break;
case 'd':
next = origin + wid;
break;
}
if (map[next] == 0) {//如果目标位置可以进入
if (next == target) {
int[] step = new int[2];
step[0] = direct;// 移动方向
step[1] = next;// 到达的新位置
stack.addElement(step);
return true;
}
if (!inStack(next)) {
int[] step = new int[2];
step[0] = direct;
step[1] = next;
stack.addElement(step);
if (findPath(map, next, target)) {
return true;
}
}
}
return false;
}

private boolean inStack(int posi) {
int[] temp;
for (int i = stack.size() - 1; i >= 0; i--) {
temp = (int[]) stack.elementAt(i);
if (posi == temp[1]) {
return true;
}
}
return false;
}
}