连连看java带图片代码,java连连看完整源代码
求java小游戏源代码
表1. CheckerDrag.java
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// CheckerDrag.javaimport java.awt.*;import java.awt.event.*;public class CheckerDrag extends java.applet.Applet{ // Dimension of checkerboard square. // 棋盘上每个小方格的尺寸 final static int SQUAREDIM = 40; // Dimension of checkerboard -- includes black outline. // 棋盘的尺寸 – 包括黑色的轮廓线 final static int BOARDDIM = 8 * SQUAREDIM + 2; // Dimension of checker -- 3/4 the dimension of a square. // 棋子的尺寸 – 方格尺寸的3/4 final static int CHECKERDIM = 3 * SQUAREDIM / 4; // Square colors are dark green or white. // 方格的颜色为深绿色或者白色 final static Color darkGreen = new Color (0, 128, 0); // Dragging flag -- set to true when user presses mouse button over checker // and cleared to false when user releases mouse button. // 拖动标记 --当用户在棋子上按下鼠标按键时设为true, // 释放鼠标按键时设为false boolean inDrag = false; // Left coordinate of checkerboard's upper-left corner. // 棋盘左上角的左方向坐标 int boardx; // Top coordinate of checkerboard's upper-left corner. //棋盘左上角的上方向坐标 int boardy; // Left coordinate of checker rectangle origin (upper-left corner). // 棋子矩形原点(左上角)的左方向坐标 int ox; // Top coordinate of checker rectangle origin (upper-left corner). // 棋子矩形原点(左上角)的上方向坐标 int oy; // Left displacement between mouse coordinates at time of press and checker // rectangle origin. // 在按键时的鼠标坐标与棋子矩形原点之间的左方向位移 int relx; // Top displacement between mouse coordinates at time of press and checker // rectangle origin. // 在按键时的鼠标坐标与棋子矩形原点之间的上方向位移 int rely; // Width of applet drawing area. // applet绘图区域的宽度 int width; // Height of applet drawing area. // applet绘图区域的高度 int height; // Image buffer. // 图像缓冲 Image imBuffer; // Graphics context associated with image buffer. // 图像缓冲相关联的图形背景 Graphics imG; public void init () { // Obtain the size of the applet's drawing area. // 获取applet绘图区域的尺寸 width = getSize ().width; height = getSize ().height; // Create image buffer. // 创建图像缓冲 imBuffer = createImage (width, height); // Retrieve graphics context associated with image buffer. // 取出图像缓冲相关联的图形背景 imG = imBuffer.getGraphics (); // Initialize checkerboard's origin, so that board is centered. // 初始化棋盘的原点,使棋盘在屏幕上居中 boardx = (width - BOARDDIM) / 2 + 1; boardy = (height - BOARDDIM) / 2 + 1; // Initialize checker's rectangle's starting origin so that checker is // centered in the square located in the top row and second column from // the left. // 初始化棋子矩形的起始原点,使得棋子在第一行左数第二列的方格里居中 ox = boardx + SQUAREDIM + (SQUAREDIM - CHECKERDIM) / 2 + 1; oy = boardy + (SQUAREDIM - CHECKERDIM) / 2 + 1; // Attach a mouse listener to the applet. That listener listens for // mouse-button press and mouse-button release events. // 向applet添加一个用来监听鼠标按键的按下和释放事件的鼠标监听器 addMouseListener (new MouseAdapter () { public void mousePressed (MouseEvent e) { // Obtain mouse coordinates at time of press. // 获取按键时的鼠标坐标 int x = e.getX (); int y = e.getY (); // If mouse is over draggable checker at time // of press (i.e., contains (x, y) returns // true), save distance between current mouse // coordinates and draggable checker origin // (which will always be positive) and set drag // flag to true (to indicate drag in progress). // 在按键时如果鼠标位于可拖动的棋子上方 // (也就是contains (x, y)返回true),则保存当前 // 鼠标坐标与棋子的原点之间的距离(始终为正值)并且 // 将拖动标志设为true(用来表明正处在拖动过程中) if (contains (x, y)) { relx = x - ox; rely = y - oy; inDrag = true; } } boolean contains (int x, int y) { // Calculate center of draggable checker. // 计算棋子的中心位置 int cox = ox + CHECKERDIM / 2; int coy = oy + CHECKERDIM / 2; // Return true if (x, y) locates with bounds // of draggable checker. CHECKERDIM / 2 is the // radius. // 如果(x, y)仍处于棋子范围内则返回true // CHECKERDIM / 2为半径 return (cox - x) * (cox - x) + (coy - y) * (coy - y) CHECKERDIM / 2 * CHECKERDIM / 2; } public void mouseReleased (MouseEvent e) { // When mouse is released, clear inDrag (to // indicate no drag in progress) if inDrag is // already set. // 当鼠标按键被释放时,如果inDrag已经为true, // 则将其置为false(用来表明不在拖动过程中) if (inDrag) inDrag = false; } }); // Attach a mouse motion listener to the applet. That listener listens // for mouse drag events. //向applet添加一个用来监听鼠标拖动事件的鼠标运动监听器 addMouseMotionListener (new MouseMotionAdapter () { public void mouseDragged (MouseEvent e) { if (inDrag) { // Calculate draggable checker's new // origin (the upper-left corner of // the checker rectangle). // 计算棋子新的原点(棋子矩形的左上角) int tmpox = e.getX () - relx; int tmpoy = e.getY () - rely; // If the checker is not being moved // (at least partly) off board, // assign the previously calculated // origin (tmpox, tmpoy) as the // permanent origin (ox, oy), and // redraw the display area (with the // draggable checker at the new // coordinates). // 如果棋子(至少是棋子的一部分)没有被 // 移出棋盘,则将之前计算的原点 // (tmpox, tmpoy)赋值给永久性的原点(ox, oy), // 并且刷新显示区域(此时的棋子已经位于新坐标上) if (tmpox boardx tmpoy boardy tmpox + CHECKERDIM boardx + BOARDDIM tmpoy + CHECKERDIM boardy + BOARDDIM) { ox = tmpox; oy = tmpoy; repaint (); } } } }); } public void paint (Graphics g) { // Paint the checkerboard over which the checker will be dragged. // 在棋子将要被拖动的位置上绘制棋盘 paintCheckerBoard (imG, boardx, boardy); // Paint the checker that will be dragged. // 绘制即将被拖动的棋子 paintChecker (imG, ox, oy); // Draw contents of image buffer. // 绘制图像缓冲的内容 g.drawImage (imBuffer, 0, 0, this); } void paintChecker (Graphics g, int x, int y) { // Set checker shadow color. // 设置棋子阴影的颜色 g.setColor (Color.black); // Paint checker shadow. // 绘制棋子的阴影 g.fillOval (x, y, CHECKERDIM, CHECKERDIM); // Set checker color. // 设置棋子颜色 g.setColor (Color.red); // Paint checker. // 绘制棋子 g.fillOval (x, y, CHECKERDIM - CHECKERDIM / 13, CHECKERDIM - CHECKERDIM / 13); } void paintCheckerBoard (Graphics g, int x, int y) { // Paint checkerboard outline. // 绘制棋盘轮廓线 g.setColor (Color.black); g.drawRect (x, y, 8 * SQUAREDIM + 1, 8 * SQUAREDIM + 1); // Paint checkerboard. // 绘制棋盘 for (int row = 0; row 8; row++) { g.setColor (((row 1) != 0) ? darkGreen : Color.white); for (int col = 0; col 8; col++) { g.fillRect (x + 1 + col * SQUAREDIM, y + 1 + row * SQUAREDIM, SQUAREDIM, SQUAREDIM); g.setColor ((g.getColor () == darkGreen) ? Color.white : darkGreen); } } } // The AWT invokes the update() method in response to the repaint() method // calls that are made as a checker is dragged. The default implementation // of this method, which is inherited from the Container class, clears the // applet's drawing area to the background color prior to calling paint(). // This clearing followed by drawing causes flicker. CheckerDrag overrides // update() to prevent the background from being cleared, which eliminates // the flicker. // AWT调用了update()方法来响应拖动棋子时所调用的repaint()方法。该方法从 // Container类继承的默认实现会在调用paint()之前,将applet的绘图区域清除 // 为背景色,这种绘制之后的清除就导致了闪烁。CheckerDrag重写了update()来 // 防止背景被清除,从而消除了闪烁。 public void update (Graphics g) { paint (g); }}
连连看JAVA源代码
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
public class lianliankan implements ActionListener
{
JFrame mainFrame; //主面板
Container thisContainer;
JPanel centerPanel,southPanel,northPanel; //子面板
JButton diamondsButton[][] = new JButton[6][5];//游戏按钮数组
JButton exitButton,resetButton,newlyButton; //退出,重列,重新开始按钮
JLabel fractionLable=new JLabel("0"); //分数标签
JButton firstButton,secondButton; //分别记录两次被选中的按钮
int grid[][] = new int[8][7];//储存游戏按钮位置
static boolean pressInformation=false; //判断是否有按钮被选中
int x0=0,y0=0,x=0,y=0,fristMsg=0,secondMsg=0,validateLV; //游戏按钮的位置坐标
int i,j,k,n;//消除方法控制
public void init(){
mainFrame=new JFrame("JKJ连连看");
thisContainer = mainFrame.getContentPane();
thisContainer.setLayout(new BorderLayout());
centerPanel=new JPanel();
southPanel=new JPanel();
northPanel=new JPanel();
thisContainer.add(centerPanel,"Center");
thisContainer.add(southPanel,"South");
thisContainer.add(northPanel,"North");
centerPanel.setLayout(new GridLayout(6,5));
for(int cols = 0;cols 6;cols++){
for(int rows = 0;rows 5;rows++ ){
diamondsButton[cols][rows]=new JButton(String.valueOf(grid[cols+1][rows+1]));
diamondsButton[cols][rows].addActionListener(this);
centerPanel.add(diamondsButton[cols][rows]);
}
}
exitButton=new JButton("退出");
exitButton.addActionListener(this);
resetButton=new JButton("重列");
resetButton.addActionListener(this);
newlyButton=new JButton("再来一局");
newlyButton.addActionListener(this);
southPanel.add(exitButton);
southPanel.add(resetButton);
southPanel.add(newlyButton);
fractionLable.setText(String.valueOf(Integer.parseInt(fractionLable.getText())));
northPanel.add(fractionLable);
mainFrame.setBounds(280,100,500,450);
mainFrame.setVisible(true);
}
public void randomBuild() {
int randoms,cols,rows;
for(int twins=1;twins=15;twins++) {
randoms=(int)(Math.random()*25+1);
for(int alike=1;alike=2;alike++) {
cols=(int)(Math.random()*6+1);
rows=(int)(Math.random()*5+1);
while(grid[cols][rows]!=0) {
cols=(int)(Math.random()*6+1);
rows=(int)(Math.random()*5+1);
}
this.grid[cols][rows]=randoms;
}
}
}
public void fraction(){
fractionLable.setText(String.valueOf(Integer.parseInt(fractionLable.getText())+100));
}
public void reload() {
int save[] = new int[30];
int n=0,cols,rows;
int grid[][]= new int[8][7];
for(int i=0;i=6;i++) {
for(int j=0;j=5;j++) {
if(this.grid[i][j]!=0) {
save[n]=this.grid[i][j];
n++;
}
}
}
n=n-1;
this.grid=grid;
while(n=0) {
cols=(int)(Math.random()*6+1);
rows=(int)(Math.random()*5+1);
while(grid[cols][rows]!=0) {
cols=(int)(Math.random()*6+1);
rows=(int)(Math.random()*5+1);
}
this.grid[cols][rows]=save[n];
n--;
}
mainFrame.setVisible(false);
pressInformation=false; //这里一定要将按钮点击信息归为初始
init();
for(int i = 0;i 6;i++){
for(int j = 0;j 5;j++ ){
if(grid[i+1][j+1]==0)
diamondsButton[i][j].setVisible(false);
}
}
}
public void estimateEven(int placeX,int placeY,JButton bz) {
if(pressInformation==false) {
x=placeX;
y=placeY;
secondMsg=grid[x][y];
secondButton=bz;
pressInformation=true;
}
else {
x0=x;
y0=y;
fristMsg=secondMsg;
firstButton=secondButton;
x=placeX;
y=placeY;
secondMsg=grid[x][y];
secondButton=bz;
if(fristMsg==secondMsg secondButton!=firstButton){
xiao();
}
}
}
public void xiao() { //相同的情况下能不能消去。仔细分析,不一条条注释
if((x0==x (y0==y+1||y0==y-1)) || ((x0==x+1||x0==x-1)(y0==y))){ //判断是否相邻
remove();
}
else{
for (j=0;j7;j++ ) {
if (grid[x0][j]==0){ //判断第一个按钮同行哪个按钮为空
if (yj) { //如果第二个按钮的Y坐标大于空按钮的Y坐标说明第一按钮在第二按钮左边
for (i=y-1;i=j;i-- ){ //判断第二按钮左侧直到第一按钮中间有没有按钮
if (grid[x][i]!=0) {
k=0;
break;
}
else //K=1说明通过了第一次验证
}
if (k==1) {
linePassOne();
}
}
if (yj){ //如果第二个按钮的Y坐标小于空按钮的Y坐标说明第一按钮在第二按钮右边
for (i=y+1;i=j ;i++ ){ //判断第二按钮左侧直到第一按钮中间有没有按钮
if (grid[x][i]!=0){
k=0;
break;
}
else
}
if (k==1){
linePassOne();
}
}
if (y==j ) {
linePassOne();
}
}
if (k==2) {
if (x0==x) {
remove();
}
if (x0x) {
for (n=x0;n=x-1;n++ ) {
if (grid[n][j]!=0) {
k=0;
break;
}
if(grid[n][j]==0 n==x-1) {
remove();
}
}
}
if (x0x) {
for (n=x0;n=x+1 ;n-- ) {
if (grid[n][j]!=0) {
k=0;
break;
}
if(grid[n][j]==0 n==x+1) {
remove();
}
}
}
}
}
for (i=0;i8;i++ ) { //列
if (grid[i][y0]==0) {
if (xi) {
for (j=x-1;j=i ;j-- ) {
if (grid[j][y]!=0) {
k=0;
break;
}
else
}
if (k==1) {
rowPassOne();
}
}
if (xi) {
for (j=x+1;j=i;j++ ) {
if (grid[j][y]!=0) {
k=0;
break;
}
else
}
if (k==1) {
rowPassOne();
}
}
if (x==i) {
rowPassOne();
}
}
if (k==2){
if (y0==y) {
remove();
}
if (y0y) {
for (n=y0;n=y-1 ;n++ ) {
if (grid[i][n]!=0) {
k=0;
break;
}
if(grid[i][n]==0 n==y-1) {
remove();
}
}
}
if (y0y) {
for (n=y0;n=y+1 ;n--) {
if (grid[i][n]!=0) {
k=0;
break;
}
if(grid[i][n]==0 n==y+1) {
remove();
}
}
}
}
}
}
}
public void linePassOne(){
if (y0j){ //第一按钮同行空按钮在左边
for (i=y0-1;i=j ;i-- ){ //判断第一按钮同左侧空按钮之间有没按钮
if (grid[x0][i]!=0) {
k=0;
break;
}
else //K=2说明通过了第二次验证
}
}
if (y0j){ //第一按钮同行空按钮在与第二按钮之间
for (i=y0+1;i=j ;i++){
if (grid[x0][i]!=0) {
k=0;
break;
}
else
}
}
}
public void rowPassOne(){
if (x0i) {
for (j=x0-1;j=i ;j-- ) {
if (grid[j][y0]!=0) {
k=0;
break;
}
else
}
}
if (x0i) {
for (j=x0+1;j=i ;j++ ) {
if (grid[j][y0]!=0) {
k=0;
break;
}
else
}
}
}
public void remove(){
firstButton.setVisible(false);
secondButton.setVisible(false);
fraction();
pressInformation=false;
k=0;
grid[x0][y0]=0;
grid[x][y]=0;
}
public void actionPerformed(ActionEvent e) {
if(e.getSource()==newlyButton){
int grid[][] = new int[8][7];
this.grid = grid;
randomBuild();
mainFrame.setVisible(false);
pressInformation=false;
init();
}
if(e.getSource()==exitButton)
System.exit(0);
if(e.getSource()==resetButton)
reload();
for(int cols = 0;cols 6;cols++){
for(int rows = 0;rows 5;rows++ ){
if(e.getSource()==diamondsButton[cols][rows])
estimateEven(cols+1,rows+1,diamondsButton[cols][rows]);
}
}
}
public static void main(String[] args) {
lianliankan llk = new lianliankan();
llk.randomBuild();
llk.init();
}
}
//old 998 lines
//new 318 lines
我用Java编写的连连看。运行后全是按钮。怎样导入图片呢。求导入图片的完整代码。在线等。
很简单,你用的Button是JButton? 那参考如下代码。
Icon icon = new ImageIcon(new URL(""));
JButton jb = new JButton(icon);
JAVA怎么实现连连看游戏图片载入的
将数字(int或Integer对象)与图片(Image对象)的对应关系保存在一个map中,当得到某个数字是,就可以得到所对应的图片了,然后在button上setIcon就可以。
求JAVA连连看游戏源代码,打包的,能运行的
123456789101112131415161718192021222324252627282930 import java.io.File;import java.util.Scanner; public class FileTest { /** * @param args */ public static void main(String[] args) { Scanner sc = new Scanner(System.in); String fileName = sc.next(); File file = new File(fileName); if (file.exists()) { if (file.isDirectory()) { String[] list = file.list(); System.out.println(list.length); for (String string : list) { System.out .println(string .substring(string.lastIndexOf(".") = 0 ? string .lastIndexOf(".") : 0)); } } else { System.out.println("不是文件夹"); } } else { System.out.println("文件不存在"); } }
用java语言编写连连看游戏
我以前自己写一个玩的。没有把你要求的功能全部实现,不过你看了后可以改一下就好了。参考一下吧,我给了注解:
package mybase.programe;
/*
* lianliankan总体算法思路:由两个确定的按钮。若这两个按钮的数字相等,就开始找它们相连的路经。这个找路经
* 分3种情况:(从下面的这三种情况,我们可以知道,需要三个检测,这三个检测分别检测一条直路经。这样就会有
* 三条路经。若这三条路经上都是空按钮,那么就刚好是三种直线(两个转弯点)把两个按钮连接起来了)
* 1.相邻
*
* 2. 若不相邻的先在第一个按钮的同行找一个空按钮。1).找到后看第二个按钮横向到这个空按钮
* 所在的列是否有按钮。2).没有的话再看第一个按钮到与它同行的那个空按钮之间是否有按钮。3).没有的话,再从
* 与第一个按钮同行的那个空按钮竖向到与第二个按钮的同行看是否有按钮。没有的话路经就通了,可以消了.
*
* 3.若2失败后,再在第一个按钮的同列找一个空按钮。1).找到后看第二个按钮竖向到这个空按钮所在的行是否有按钮。
* 2).没有的话,再看第一个按钮到与它同列的那个空按钮之间是否有按钮。3).没有的话,再从与第一个按钮同列的
* 那个空按钮横向到与第二个按钮同列看是否有按钮。没有的话路经就通了,可以消了。
*
* 若以上三步都失败,说明这两个按钮不可以消去。
*/
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
public class LianLianKan implements ActionListener {
JFrame mainFrame; // 主面板
Container thisContainer;
JPanel centerPanel, southPanel, northPanel; // 子面板
JButton diamondsButton[][] = new JButton[6][5];// 游戏按钮数组
JButton exitButton, resetButton, newlyButton; // 退出,重列,重新开始按钮
JLabel fractionLable = new JLabel("0"); // 分数标签
JButton firstButton, secondButton; // 分别记录两次被选中的按钮
// 储存游戏按钮位置(这里其实只要6行,5列。但是我们用了8行,7列。是等于在这个面板按钮的周围还围
//了一层是0的按钮,这样就可以实现靠近面板边缘的两个按钮可以消去)
int grid[][] = new int[8][7];
static boolean pressInformation = false; // 判断是否有按钮被选中
int x0 = 0, y0 = 0, x = 0, y = 0, fristMsg = 0, secondMsg = 0, validateLV; // 游戏按钮的位置坐标
int i, j, k, n;// 消除方法控制
public void init() {
mainFrame = new JFrame("JKJ连连看");
thisContainer = mainFrame.getContentPane();
thisContainer.setLayout(new BorderLayout());
centerPanel = new JPanel();
southPanel = new JPanel();
northPanel = new JPanel();
thisContainer.add(centerPanel, "Center");
thisContainer.add(southPanel, "South");
thisContainer.add(northPanel, "North");
centerPanel.setLayout(new GridLayout(6, 5));
for (int cols = 0; cols 6; cols++) {
for (int rows = 0; rows 5; rows++) {
diamondsButton[cols][rows] = new JButton(String
.valueOf(grid[cols + 1][rows + 1]));
diamondsButton[cols][rows].addActionListener(this);
centerPanel.add(diamondsButton[cols][rows]);
}
}
exitButton = new JButton("退出");
exitButton.addActionListener(this);
resetButton = new JButton("重列");
resetButton.addActionListener(this);
newlyButton = new JButton("再来一局");
newlyButton.addActionListener(this);
southPanel.add(exitButton);
southPanel.add(resetButton);
southPanel.add(newlyButton);
fractionLable.setText(String.valueOf(Integer.parseInt(fractionLable
.getText())));
northPanel.add(fractionLable);
mainFrame.setBounds(280, 100, 500, 450);
mainFrame.setVisible(true);
mainFrame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
}
public void randomBuild() {
int randoms, cols, rows;
for (int twins = 1; twins = 15; twins++) {//一共15对button,30个
randoms = (int) (Math.random() * 25 + 1);//button上的数字
for (int alike = 1; alike = 2; alike++) {
cols = (int) (Math.random() * 6 + 1);
rows = (int) (Math.random() * 5 + 1);
while (grid[cols][rows] != 0) {//等于0说明这个空格有了button
cols = (int) (Math.random() * 6 + 1);
rows = (int) (Math.random() * 5 + 1);
}
this.grid[cols][rows] = randoms;
}
}
}
public void fraction() {
fractionLable.setText(String.valueOf(Integer.parseInt(fractionLable
.getText()) + 100));
}
public void reload() {
int save[] = new int[30];
int n = 0, cols, rows;
int grid[][] = new int[8][7];
for (int i = 0; i = 6; i++) {
for (int j = 0; j = 5; j++) {
if (this.grid[i][j] != 0) {
save[n] = this.grid[i][j];//记下每个button的数字
n++;//有几个没有消去的button
}
}
}
n = n - 1;
this.grid = grid;
while (n = 0) {//把没有消去的button重新放一次
cols = (int) (Math.random() * 6 + 1);
rows = (int) (Math.random() * 5 + 1);
while (grid[cols][rows] != 0) {
cols = (int) (Math.random() * 6 + 1);
rows = (int) (Math.random() * 5 + 1);
}
this.grid[cols][rows] = save[n];
n--;
}
mainFrame.setVisible(false);
pressInformation = false; // 这里一定要将按钮点击信息归为初始
init();
for (int i = 0; i 6; i++) {
for (int j = 0; j 5; j++) {
if (grid[i + 1][j + 1] == 0)
diamondsButton[i][j].setVisible(false);
}
}
}
public void estimateEven(int placeX, int placeY, JButton bz) {
if (pressInformation == false) {
x = placeX;
y = placeY;
secondMsg = grid[x][y];
secondButton = bz;
pressInformation = true;
} else {
x0 = x;
y0 = y;
fristMsg = secondMsg;
firstButton = secondButton;
x = placeX;
y = placeY;
secondMsg = grid[x][y];
secondButton = bz;
if (fristMsg == secondMsg secondButton != firstButton) {
xiao();
}
}
}
public void xiao() { // 相同的情况下能不能消去。仔细分析,不一条条注释
if ((x0 == x (y0 == y + 1 || y0 == y - 1))
|| ((x0 == x + 1 || x0 == x - 1) (y0 == y))) { // 判断是否相邻
remove();
} else {
for (j = 0; j 7; j++) {
if (grid[x0][j] == 0) { // 判断和第一个按钮同行的哪个按钮为空
//如果找到一个为空的,就按列值的三种情况比较第二个按钮与空按钮的位置
if (y j) {//第二个按钮在空按钮右边
for (i = y - 1; i = j; i--) { //检测从第二个按钮横向左边到空格所在列为止是否全是空格
if (grid[x][i] != 0) {
k = 0;
break;//存在非空格的就退出,这一退出就不可能k==2了,所以就会到下而215行出同理的判断列
} else {
k = 1;
} // K=1说明全是空格通过了第一次验证,也就是从第二个按钮横向左边到空格所在列为止全是空格
}
if (k == 1) {
linePassOne();//进入第二次验证,也就是从第一个按钮到它同行的空格之间的空格判断
}
}
if (y j) { // 第二个按钮在空按钮左边
for (i = y + 1; i = j; i++) {//检测从第二个按钮横向右边到空格所在列为止是否全是空格
if (grid[x][i] != 0) {
k = 0;
break;
} else {
k = 1;
}
}
if (k == 1) {
linePassOne();
}
}
if (y == j) {//第二个按钮和空按钮同列
linePassOne();
}
}
//第三次检测,检测确定为空的第j列的那个按钮竖向到第二个按钮,看是不是有按钮
if (k == 2) {
if (x0 == x) {//第一,二按钮在同行
remove();
}
if (x0 x) {//第一按钮在第二按钮下边
for (n = x0; n = x - 1; n++) {//从空按钮竖向到第二个按钮所在行是否有按钮
if (grid[n][j] != 0) {
k = 0;
break;
}
//没有按钮,说明这条路经就通了
if (grid[n][j] == 0 n == x - 1) {
remove();
}
}
}
if (x0 x) {//第一按钮在第二按钮上边
for (n = x0; n = x + 1; n--) {
if (grid[n][j] != 0) {
k = 0;
break;
}
if (grid[n][j] == 0 n == x + 1) {
remove();
}
}
}
}
}//-------------------------------------for
//当上面的检测与第一个按钮同行的空格按钮失败后(不能找到与第二个按钮的相连路经),下面就执行
//检测与第一个按钮同列的空格按钮
for (i = 0; i 8; i++) {
if (grid[i][y0] == 0) {// 判断和第一个按钮同列的哪个按钮为空
if (x i) {//第二个按钮在这个空按钮的下面
for (j = x - 1; j = i; j--) {
if (grid[j][y] != 0) {
k = 0;
break;
} else {
k = 1;
}
}
if (k == 1) {
rowPassOne();
}
}
if (x i) {//第二个按钮在这个空按钮的上面
for (j = x + 1; j = i; j++) {
if (grid[j][y] != 0) {
k = 0;
break;
} else {
k = 1;
}
}
if (k == 1) {
rowPassOne();
}
}
if (x == i) {//第二个按钮与这个空按钮同行
rowPassOne();
}
}
if (k == 2) {
if (y0 == y) {//第二个按钮与第一个按钮同列
remove();
}
if (y0 y) {//第二个按钮在第一个按钮右边
for (n = y0; n = y - 1; n++) {
if (grid[i][n] != 0) {
k = 0;
break;
}
if (grid[i][n] == 0 n == y - 1) {
remove();
}
}
}
if (y0 y) {//第二个按钮在第一个按钮左边
for (n = y0; n = y + 1; n--) {
if (grid[i][n] != 0) {
k = 0;
break;
}
if (grid[i][n] == 0 n == y + 1) {
remove();
}
}
}
}
}//--------------------------------for
}//-------------else
}//------------xiao
public void linePassOne() {
if (y0 j) { // 第一按钮同行空按钮在左边
for (i = y0 - 1; i = j; i--) { // 判断第一按钮同左侧空按钮之间有没按钮
if (grid[x0][i] != 0) {
k = 0;
break;
} else {
k = 2;
} // K=2说明通过了第二次验证
}
}
if (y0 j) { // 第一按钮同行空按钮在右边
for (i = y0 + 1; i = j; i++) {
if (grid[x0][i] != 0) {
k = 0;
break;
} else {
k = 2;
}
}
}
}
public void rowPassOne() {
if (x0 i) {//第一个按钮在与它同列的那个空格按钮下面
for (j = x0 - 1; j = i; j--) {
if (grid[j][y0] != 0) {
k = 0;
break;
} else {
k = 2;
}
}
}
if (x0 i) {//第一个按钮在与它同列的那个空格按钮上面
for (j = x0 + 1; j = i; j++) {
if (grid[j][y0] != 0) {
k = 0;
break;
} else {
k = 2;
}
}
}
}
public void remove() {
firstButton.setVisible(false);
secondButton.setVisible(false);
fraction();
pressInformation = false;
k = 0;
grid[x0][y0] = 0;
grid[x][y] = 0;
}
public void actionPerformed(ActionEvent e) {
if (e.getSource() == newlyButton) {
int grid[][] = new int[8][7];
this.grid = grid;
randomBuild();
mainFrame.setVisible(false);
pressInformation = false;
init();
}
if (e.getSource() == exitButton)
System.exit(0);
if (e.getSource() == resetButton)
reload();
for (int cols = 0; cols 6; cols++) {
for (int rows = 0; rows 5; rows++) {
if (e.getSource() == diamondsButton[cols][rows])
estimateEven(cols + 1, rows + 1, diamondsButton[cols][rows]);
}
}
}
public static void main(String[] args) {
LianLianKan llk = new LianLianKan();
llk.randomBuild();
llk.init();
}
}
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