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请帮忙用数据结构（java版）的知识解决这道迷宫问题的程序代码。
我这是用c写的。你可以看看，希望能帮助到你。
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#include"stdlib.h"
#include"stdio.h"
#define N 50
#define M 50
int X;
int maze[N+2][M+2];
struct point{
int row,col,predecessor;
}queue[512];
int head=0,tail=0;
void shoudong_maze(int m,int n){
int i,j;
printf("\n\n");
printf("请按行输入迷宫，0表示通路，1表示障碍:\n\n");
for(i=0;im;i++)
for(j=0;jn;j++)
scanf("%d",maze[i][j]);
}
void zidong_maze(int m,int n){
int i,j;
printf("\n迷宫生成中……\n\n");
system("pause");
for(i=0;im;i++)
for(j=0;jn;j++)
maze[i][j]=rand()%2;
//由于rand()产生的随机数是从0到RAND_MAX
//RAND_MAX是定义在stdlib.h中的,其值至少为32767)
//要产生从X到Y的数,只需要这样写：k=rand()%(Y-X+1)+X;
}
void print_maze(int m,int n){
int i,j;
printf("\n迷宫生成结果如下:\n\n");
printf("迷宫入口\n");
printf("↓");
for(i=0;im;i++)
{printf("\n");
for(j=0;jn;j++)
{if(maze[i][j]==0) printf("□");
if(maze[i][j]==1) printf("■");}
}
printf("→迷宫出口\n");
}
void result_maze(int m,int n)
{ int i,j;
printf("迷宫通路(用☆表示)如下所示：\n\t");
for(i=0;im;i++)
{ printf("\n");
for(j=0;jn;j++)
{if(maze[i][j]==0||maze[i][j]==2) printf("□");
if(maze[i][j]==1) printf("■");
if(maze[i][j]==3) printf("☆");
}
}
}
void enqueue(struct point p)
{ queue[tail]=p;
tail++;
}
struct point dequeue()
{ head++;
return queue[head-1];
}
int is_empty()
{ return head==tail;
}
void visit(int row,int col,int maze[52][52])
{ struct point visit_point={row,col,head-1};
maze[row][col]=2;
enqueue(visit_point);
}
int mgpath(int maze[52][52],int m,int n)
{ X=1;
struct point p={0,0,-1};
if(maze[p.row][p.col]==1)
{ printf("\n===============================================\n");
printf("此迷宫无解\n\n");X=0;return 0;}
maze[p.row][p.col]=2;
enqueue(p);
while(!is_empty())
{p=dequeue();
if((p.row==m-1)(p.col==n-1)) break;
if((p.col+1n)(maze[p.row][p.col+1]==0)) visit(p.row,p.col+1,maze);
if((p.row+1m)(maze[p.row+1][p.col]==0)) visit(p.row+1,p.col,maze);
if((p.col-1=0)(maze[p.row][p.col-1]==0)) visit(p.row,p.col-1,maze);
if((p.row-1=0)(maze[p.row-1][p.col]==0)) visit(p.row-1,p.col,maze);
}
if(p.row==m-1p.col==n-1)
{printf("\n==================================================================\n");
printf("迷宫路径为：\n");
printf("(%d,%d)\n",p.row,p.col);
maze[p.row][p.col]=3;
while(p.predecessor!=-1)
{p=queue[p.predecessor];
printf("(%d,%d)\n",p.row,p.col);
maze[p.row][p.col]=3;
}
}
else {printf("\n=============================================================\n");
printf("此迷宫无解！\n\n");X=0;}
return 0;
}
int main()
{int i,m,n,cycle=0;
while(cycle!=(-1))
{
printf("********************************************************************************\n");
printf(" ☆欢迎进入迷宫求解系统☆\n");
printf(" 设计者:尹旭 林静波（信息2班）\n");
printf("********************************************************************************\n");
printf(" 手动生成迷宫 请按：1\n");
printf(" 自动生成迷宫 请按：2\n");
printf(" 退出 请按：3\n\n");
printf("********************************************************************************\n");
printf("\n");
printf("请选择你的操作：\n");
scanf("%d",i);
switch(i)
{case 1:printf("\n请输入行数：");
scanf("%d",m);
printf("\n");
printf("请输入列数：");
scanf("%d",n);
while((m=0||m50)||(n=0||n50))
{ printf("\n抱歉，你输入的行列数超出预设范围(0-50,0-50),请重新输入：\n\n");
printf("请输入行数：");
scanf("%d",m);
printf("\n");
printf("请输入列数：");
scanf("%d",n);
}
shoudong_maze(m,n);
print_maze(m,n);
mgpath(maze,m,n);
if(X!=0)
result_maze(m,n);
printf("\n\nPress Enter Contiue!\n");
getchar();
while(getchar()!='\n');
break;
case 2:printf("\n请输入行数：");
scanf("%d",m);
printf("\n");
printf("请输入列数：");
scanf("%d",n);
while((m=0||m50)||(n=0||n50))
{printf("\n抱歉，你输入的行列数超出预设范围(0-50,0-50),请重新输入：\n\n");
printf("请输入行数：");
scanf("%d",m);
printf("\n");
printf("请输入列数：");
scanf("%d",n);
}
zidong_maze(m,n);
print_maze(m,n);
mgpath(maze,m,n);
if(X!=0)
result_maze(m,n);
printf("\n\nPress Enter Contiue!\n");getchar();while(getchar()!='\n');break;
case 3:cycle=(-1);
break;
default:printf("\n");
printf("你的输入有误!\n");
printf("\nPress Enter Contiue!\n");
getchar();
while(getchar()!='\n');break;
}
}
}
关于Java走迷宫的问题。我已经有相关代码了，但是我看不懂。麻烦高手帮忙注释一下，然后再修改点儿。
package 走迷宫;
import java.awt.BorderLayout;
import java.awt.Color;
import java.awt.GridLayout;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.TimerTask;
import javax.swing.JButton;
import javax.swing.JFrame;
import javax.swing.JOptionPane;
import javax.swing.JPanel;
// 迷宫
public class Maze extends JFrame implements ActionListener {
private JPanel panel;
private JPanel northPanel;
private JPanel centerPanel;
private MazeGrid grid[][];
private JButton restart;
private JButton dostart;
private int rows;// rows 和cols目前暂定只能是奇数
private int cols;
private ListString willVisit;
private ListString visited;
private LinkedListString comed;
private long startTime;
private long endTime;
public Maze() {
rows = 25;
cols = 25;
willVisit = new ArrayListString();
visited = new ArrayListString();
comed = new LinkedListString();
init();
this.setTitle("回溯法--走迷宫");
this.add(panel);
this.pack();
this.setVisible(true);
this.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
}
public void init() {
panel = new JPanel();
northPanel = new JPanel();
centerPanel = new JPanel();
panel.setLayout(new BorderLayout());
restart = new JButton("重新生成迷宫");
dostart = new JButton("开始走迷宫");
grid = new MazeGrid[rows][cols];
centerPanel.setLayout(new GridLayout(rows, cols, 1, 1));
centerPanel.setBackground(new Color(0, 0, 0));
northPanel.add(restart);
northPanel.add(dostart);
dostart.addActionListener(this);
restart.addActionListener(this);
for (int i = 0; i grid.length; i++)
for (int j = 0; j grid[i].length; j++) {
if (j % 2 == 0 i % 2 == 0)
grid[i][j] = new MazeGrid(true, 20, 20);
else
grid[i][j] = new MazeGrid(false, 20, 20);
}
grid[0][0].setVisited(true);
grid[0][0].setPersonCome(true);
grid[0][0].setStart(true);
visited.add("0#0");
grid[rows - 1][cols - 1].setEnd(true);
grid = createMap(grid, 0, 0);
for (int i = 0; i grid.length; i++)
for (int j = 0; j grid[i].length; j++) {
grid[i][j].repaint();
centerPanel.add(grid[i][j]);
}
panel.add(northPanel, BorderLayout.NORTH);
panel.add(centerPanel, BorderLayout.CENTER);
}
/**
* 生成迷宫
*
* @param mazeGrid
* @param x
* @param y
* @return
*/
public MazeGrid[][] createMap(MazeGrid mazeGrid[][], int x, int y) {
int visitX = 0;
int visitY = 0;
if (x - 2 = 0) {
if (!mazeGrid[x - 2][y].isVisited()) {
willVisit.add((x - 2) + "#" + y);
}
}
if (x + 2 cols) {
if (!mazeGrid[x + 2][y].isVisited()) {
willVisit.add((x + 2) + "#" + y);
}
}
if (y - 2 = 0) {
if (!mazeGrid[x][y - 2].isVisited()) {
willVisit.add(x + "#" + (y - 2));
}
}
if (y + 2 rows) {
if (!mazeGrid[x][y + 2].isVisited()) {
willVisit.add(x + "#" + (y + 2));
}
}
if (!willVisit.isEmpty()) {
int visit = (int) (Math.random() * willVisit.size());
String id = willVisit.get(visit);
visitX = Integer.parseInt(id.split("#")[0]);
visitY = Integer.parseInt(id.split("#")[1]);
mazeGrid[(visitX + x) / 2][(visitY + y) / 2].setMark(true);
mazeGrid[visitX][visitY].setVisited(true);
if (!visited.contains(id)) {// 将这个点加到已访问中去
visited.add(id);
}
willVisit.clear();
createMap(mazeGrid, visitX, visitY);
} else {
if (!visited.isEmpty()) {
String id = visited.remove(visited.size() - 1);// 取出最后一个元素
visitX = Integer.parseInt(id.split("#")[0]);
visitY = Integer.parseInt(id.split("#")[1]);
mazeGrid[visitX][visitY].setVisited(true);
createMap(mazeGrid, visitX, visitY);
}
}
return mazeGrid;
}
/**
* 走迷宫
*
* @param mazeGrid
* @param x
* @param y
*/
public String goMaze(MazeGrid mazeGrid[][], int x, int y) {
int comeX = 0;
int comeY = 0;
// left
if (x - 1 = 0) {
if (mazeGrid[x - 1][y].isMark()) {
if (!comed.contains((x - 1) + "#" + y))
willVisit.add((x - 1) + "#" + y);
}
}
// right
if (x + 1 cols) {
if (mazeGrid[x + 1][y].isMark()) {
if (!comed.contains((x + 1) + "#" + y))
willVisit.add((x + 1) + "#" + y);
}
}
// up
if (y - 1 = 0) {
if (mazeGrid[x][y - 1].isMark()) {
if (!comed.contains(x + "#" + (y - 1)))
willVisit.add(x + "#" + (y - 1));
}
}
// down
if (y + 1 rows) {
if (mazeGrid[x][y + 1].isMark()) {
if (!comed.contains(x + "#" + (y + 1)))
willVisit.add(x + "#" + (y + 1));
}
}
if (!willVisit.isEmpty()) {
int visit = (int) (Math.random() * willVisit.size());
String id = willVisit.get(visit);
comeX = Integer.parseInt(id.split("#")[0]);
comeY = Integer.parseInt(id.split("#")[1]);
mazeGrid[x][y].setPersonCome(false);
mazeGrid[comeX][comeY].setPersonCome(true);
mazeGrid[x][y].repaint();
mazeGrid[comeX][comeY].repaint();
willVisit.clear();
comed.add(x + "#" + y);
} else {
if (!comed.isEmpty()) {
String id = comed.removeLast();
comeX = Integer.parseInt(id.split("#")[0]);
comeY = Integer.parseInt(id.split("#")[1]);
mazeGrid[x][y].setPersonCome(false);
mazeGrid[comeX][comeY].setPersonCome(true);
mazeGrid[x][y].repaint();
mazeGrid[comeX][comeY].repaint();
comed.addFirst(x + "#" + y);
}
}
return comeX + "#" + comeY;
}
int comeX = 0;
int comeY = 0;
public void actionPerformed(ActionEvent e) {
if (e.getActionCommand().equals("重新生成迷宫")) {
refreshMap(grid);
} else if (e.getActionCommand().equals("开始走迷宫")) {
startTime = System.currentTimeMillis();
dostart.setVisible(false);
restart.setText("禁止刷新");
int delay = 1000;
int period = 500;// 循环间隔
java.util.Timer timer = new java.util.Timer();
timer.scheduleAtFixedRate(new TimerTask() {
public void run() {
if (grid[rows - 1][cols - 1].isPersonCome()) {
endTime = System.currentTimeMillis();
JOptionPane.showMessageDialog(null, "已经走出迷宫，耗时"
+ (endTime - startTime) / 1000 + "秒", "消息提示",
JOptionPane.ERROR_MESSAGE);
this.cancel();
restart.setText("重新生成迷宫");
} else {
String id = goMaze(grid, comeX, comeY);
comeX = Integer.parseInt(id.split("#")[0]);
comeY = Integer.parseInt(id.split("#")[1]);
}
}
}, delay, period);
}
}
/**
* 刷新地图
*/
public void refreshMap(MazeGrid mazeGrid[][]) {
comeX = 0;
comeY = 0;
willVisit.clear();
visited.clear();
comed.clear();
this.remove(panel);
init();
this.add(panel);
this.pack();
this.setVisible(true);
}
public static void main(String args[]) {
long start = System.currentTimeMillis();
new Maze();
long end = System.currentTimeMillis();
System.out.println("使用ArrayList生成迷宫耗时：" + (end - start) + "毫秒");
}
}
用java语言控制小机器人走迷宫的算法?
我昨天刚写了个走迷宫的界面(一个初始小球，一个目标小球，随机在界面种生成障碍(迷宫图)，然后初始小球移动到目标小球那)，不知道是否跟你的想法一样。用的是回溯法(目前我只知道这个算法走迷宫)，你可以查下。PS:我电脑没联网不能把代码给你…QQ254774042。
求用java语言寻找走出迷宫路线的算法
首先给定一个初始坐标，然后构建一个容器保存坐标值，之后进行迭代，横坐标+1，或者纵坐标+1，这个顺寻自己定义（四个方向），经过的“路径”保存在那个容器中，如果遇到死角，以此往回迭代，在容器中将遇到死角的那个坐标删除。最后找到自己定义的那个迷宫出口坐标。
JAVA走迷宫小游戏
import java.awt.*;
import javax.swing.*;
import java.awt.event.*;
import javax.swing.event.*;
class JPanelEa1 extends JFrame implements MenuListener,ActionListener
{
JPanel mainp,p1,p2,p3,p4;
JTextField jt1;
JMenuItem mnuCopy;
JMenuItem mnuPaste;
JButton bM;
boolean isDouble = false;// 是否为实数
int opFlag = -1;
static double t1 = 0, t2 = 0, t3 = 0, result = 0;
static int opflag1 = -1, opflag2 = -1, flag = 0, resflag = 1;
int preOp, currentOp = 0;// 标准位
double op1 = 0, op2 = 0;// 操作数
double n3;
StringBuffer buf = new StringBuffer(20);
StringBuffer copyBoard = new StringBuffer(20);// 剪贴板
StringBuffer memory = new StringBuffer(20);// M系列
StringBuffer str = new StringBuffer();
public JPanelEa1()
{
p1=new JPanel();
p1.setLayout(new GridLayout(2,1,10,10));
JMenuBar mnuNotepad=new JMenuBar();
JMenu mnuEdit=new JMenu("编辑(E)");
mnuEdit.setMnemonic(KeyEvent.VK_E);
JMenu mnuCheck=new JMenu("查看(V)");
mnuCheck.setMnemonic(KeyEvent.VK_V);
JMenu mnuHelp=new JMenu("帮助(H)");
mnuCopy=new JMenuItem("复制(C)");
mnuPaste=new JMenuItem("粘贴(P)");
JMenuItem mnuVisit=new JMenuItem("查看帮助(V)");
JMenuItem mnuAbout=new JMenuItem("关于计算器(A)");
JSeparator sep=new JSeparator();
jt1=new JTextField("0.");
jt1.setEnabled(false);
jt1.setHorizontalAlignment(JTextField.RIGHT);
mnuEdit.addMenuListener(this);
mnuCheck.addMenuListener(this);
mnuHelp.addMenuListener(this);
mnuCopy.addActionListener(this);
mnuPaste.addActionListener(this);
mnuVisit.addActionListener(this);
mnuAbout.addActionListener(this);
mnuNotepad.add(mnuEdit);
mnuNotepad.add(mnuCheck);
mnuNotepad.add(mnuHelp);
mnuEdit.add(mnuCopy);
mnuEdit.add(mnuPaste);
mnuEdit.add(sep);
mnuHelp.add(mnuVisit);
mnuHelp.add(sep);
mnuHelp.add(mnuAbout);
p1.add(mnuNotepad);
p1.add(jt1);
p2=new JPanel();
p2.setLayout(new FlowLayout(FlowLayout.CENTER));
bM = new JButton(" ");
bM.setEnabled(false);
p2.add(bM);
Button b25=new Button("Backspace");
b25.addActionListener(this);
Button b26=new Button("CE");
b26.addActionListener(this);
Button b27=new Button("C");
b27.addActionListener(this);
p2.add(b25);
p2.add(b26);
p2.add(b27);
p3=new JPanel();
p3.setLayout(new GridLayout(4,6));
Button button1=new Button("MC");
button1.addActionListener(this);
Button button2=new Button("7");
button2.addActionListener(this);
Button button3=new Button("8");
button3.addActionListener(this);
Button button4=new Button("9");
button4.addActionListener(this);
Button button5=new Button("/");
button5.addActionListener(this);
Button button6=new Button("sqrt");
button6.addActionListener(this);
Button button7=new Button("MR");
button7.addActionListener(this);
Button button8=new Button("4");
button8.addActionListener(this);
Button button9=new Button("5");
button9.addActionListener(this);
Button button10=new Button("6");
button10.addActionListener(this);
Button button11=new Button("*");
button11.addActionListener(this);
Button button12=new Button("%");
button12.addActionListener(this);
Button button13=new Button("MS");
button13.addActionListener(this);
Button button14=new Button("1");
button14.addActionListener(this);
Button button15=new Button("2");
button15.addActionListener(this);
Button button16=new Button("3");
button16.addActionListener(this);
Button button17=new Button("-");
button17.addActionListener(this);
Button button18=new Button("1/x");
button18.addActionListener(this);
Button button19=new Button("M+");
button19.addActionListener(this);
Button button20=new Button("0");
button20.addActionListener(this);
Button button21=new Button("+/-");
button21.addActionListener(this);
Button button22=new Button(".");
button22.addActionListener(this);
Button button23=new Button("+");
button23.addActionListener(this);
Button button24=new Button("=");
button24.addActionListener(this);
p3.add(button1);p3.add(button2);
p3.add(button3);p3.add(button4);
p3.add(button5);p3.add(button6);
p3.add(button7);p3.add(button8);
p3.add(button9);p3.add(button10);
p3.add(button11);p3.add(button12);
p3.add(button13);p3.add(button14);
p3.add(button15);p3.add(button16);
p3.add(button17);p3.add(button18);
p3.add(button19);p3.add(button20);
p3.add(button21);p3.add(button22);
p3.add(button23);p3.add(button24);
setSize(250,150);
mainp=new JPanel();
mainp.setLayout(new GridLayout(3,1));
mainp.add(p1);
mainp.add(p2);
mainp.add(p3);
setContentPane(mainp);
setTitle("计算器");
setSize(300,300);
setVisible(true);
setDefaultCloseOperation(EXIT_ON_CLOSE);
}
public void actionPerformed(ActionEvent e)
{
String s= e.getActionCommand();
if (s.equals("复制(C)")) {
String temp = jt1.getText().trim();
copyBoard.replace(0, copyBoard.length(), temp);
mnuPaste.setEnabled(true);
} else if (s.equals("粘贴(p)")) {
jt1.setText(copyBoard.toString());
} else if (s.equals("CE")) {
// 如果是CE则清除文本框
jt1.setText("0.");
} else if (s.equals("Backspace")) {
if (!jt1.getText().trim().equals("0.")) {
// 如果文本框中有内容
if (str.length() != 1 str.length() != 0) {
jt1.setText(str.delete(str.length() - 1, str.length())
.toString());
} else {
jt1.setText("0.");
str.setLength(0);
}
}
op2 = Double.parseDouble(jt1.getText().trim());
} else if (s.equals("C")) {
// 如果是C删除当前计算
jt1.setText("0.");
op1 = op2 = 0;
str.replace(0, str.length(), " ");
preOp = currentOp = 0;
} else if (s.equals("MC")) {
// 如果是MC则清除缓冲区
String temp = "";
memory.replace(0, memory.length(), temp);
bM.setText(" ");
} else if (s.equals("MR")) {
// 如果按键为MR则恢复缓冲区的数到文本框
jt1.setText(memory.toString());
} else if (s.equals("MS")) {
// 如果按键为MS则将文本框的数存入缓冲区
String s1 = jt1.getText().trim();
memory.replace(0, memory.length(), s1);
bM.setText("M");
} else if (s.equals("M+")) {
// 如果按键为MS则将文本框值与缓冲区的数相加但不显示结果
String temp1 = jt1.getText().trim();
double dtemp = Double.parseDouble(temp1);
String temp2 = memory.toString();
dtemp += Double.parseDouble(temp2);
temp1 = String.valueOf(dtemp);
memory.replace(0, memory.length(), temp1);
} else if (s.equals("1/x")) {
// 如果按键为1/x则将文本框中的数据为它的倒数
String temp = jt1.getText().trim();
double dtemp = Double.parseDouble(temp);
jt1.setText("" + 1 / dtemp);
} else if (s.equals("sqrt")) {
// 如果按键为sqrt则将文本框中的内容求平方根
String temp = jt1.getText().trim();
double dtemp = Double.parseDouble(temp);
jt1.setText("" + Math.sqrt(dtemp));
} else if (s.equals("+")) {
str.setLength(0);
if (currentOp == 0) {
preOp = currentOp = 1;
op2 = 0;
jt1.setText("" + op1);
} else {
currentOp = preOp;
preOp = 1;
switch (currentOp) {
case 1:
op1 += op2;
jt1.setText("" + op1);
break;
case 2:
op1 -= op2;
jt1.setText("" + op1);
break;
case 3:
op1 *= op2;
jt1.setText("" + op1);
break;
case 4:
op1 /= op2;
jt1.setText("" + op1);
break;
}
}
} else if (s.equals("-")) {
str.setLength(0);
if (currentOp == 0) {
preOp = currentOp = 2;// op1=op2;op2=0;
jt1.setText("" + op1);
} else {
currentOp = preOp;
preOp = 2;
switch (currentOp) {
case 1:
op1 = op1 + op2;
jt1.setText("" + op1);
break;
case 2:
op1 = op1 - op2;
jt1.setText("" + op1);
break;
case 3:
op1 = op1 * op2;
jt1.setText("" + op1);
break;
case 4:
op1 = op1 / op2;
jt1.setText("" + op1);
break;
}
}
} else if (s.equals("*"))// *
{
str.setLength(0);
if (currentOp == 0) {
preOp = currentOp = 3;// op1=op2;op2=1;
jt1.setText("" + op1);// op1=op2;
} else {
currentOp = preOp;
preOp = 3;
switch (currentOp) {
case 1:
op1 = op1 + op2;
jt1.setText("" + op1);
break;
case 2:
op1 = op1 - op2;
jt1.setText("" + op1);
break;
case 3:
op1 = op1 * op2;
jt1.setText("" + op1);
break;
case 4:
op1 = op1 / op2;
jt1.setText("" + op1);
break;
}
}
} else if (s.equals("/"))// /
{
str.setLength(0);
if (currentOp == 0) {
preOp = currentOp = 4;// op2=1;
jt1.setText("" + op1);// op1=op2;
} else {
currentOp = preOp;
preOp = 4;
switch (currentOp) {
case 1:
op1 = op1 + op2;
jt1.setText("" + op1);
break;
case 2:
op1 = op1 - op2;
jt1.setText("" + op1);
break;
case 3:
op1 = op1 * op2;
jt1.setText("" + op1);
break;
case 4:
op1 = op1 / op2;
jt1.setText("" + op1);
break;
}
}
} else if (s.equals("="))// =
{
if (currentOp == 0) {
str.setLength(0);
jt1.setText("" + op2);
} else {
str.setLength(0);
currentOp = preOp;
switch (currentOp) {
case 1:
op1 = op1 + op2;
jt1.setText("" + op1);
break;
case 2:
op1 = op1 - op2;
jt1.setText("" + op1);
break;
case 3:
op1 = op1 * op2;
jt1.setText("" + op1);
break;
case 4:
op1 = op1 / op2;
jt1.setText("" + op1);
break;
}
currentOp = 0;
op2 = 0;
}
} else if (s.equals(".")) {
isDouble = true;
if (jt1.getText().trim().indexOf('.') != -1)
;
else {
if (jt1.getText().trim().equals("0")) {
str.setLength(0);
jt1.setText((str.append("0" + s)).toString());
}
// else
// if(jt1.getText().trim().equals("")){}//如果初时显示为空则不做任何操作
else {
jt1.setText((str.append(s)).toString());
}
}
} else if (s.equals("0"))// 如果选择的是"0"这个数字键
{
if (jt1.getText().trim().equals("0.")) {
} else {
jt1.setText(str.append(s).toString());
op2 = Double.parseDouble(jt1.getText().trim());
}
} else {
jt1.setText(str.append(s).toString());
op2 = Double.parseDouble(jt1.getText().trim());
if (currentOp == 0)
op1 = op2;
}
}
public static void main(String args[])
{
JPanelEa1 g=new JPanelEa1();
}
@Override
public void menuSelected(MenuEvent e) {
// TODO Auto-generated method stub
}
@Override
public void menuDeselected(MenuEvent e) {
// TODO Auto-generated method stub
}
@Override
public void menuCanceled(MenuEvent e) {
// TODO Auto-generated method stub
}
}
java老鼠迷宫代码难吗
非常难。思路：
1、设老鼠的行进路线都是优先选择下-右-上-左。
2、设老鼠很聪明，走过的路线走撒泡尿，表示鼠大爷到此一游，我们可以把数组的值改为3，表示走过，但走不通。
3、这是一个int[8][8]的二位数组，那么开始位置下标是1,1，结束位置是6,6。行和列分别用、j表示。
4、实际路线我们可以设置2表示，我们可以使用递归，让老鼠不断测试路线。
5、最后打印数组，看老鼠的实际路线。
本文名称：java走迷宫代码怎么用 Java实现迷宫程序
转载注明：http://scyingshan.cn/article/hpohpi.html
