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求java小游戏源代码表1. CheckerDrag.java
// 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 * SQUAREDIM2;// Dimension of checker -- 3/4 the dimension of a square.// 棋子的尺寸 – 方格尺寸的3/4final 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falseboolean 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) / 21;boardy = (height - BOARDDIM) / 21;// 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 = boardxSQUAREDIM(SQUAREDIM - CHECKERDIM) / 21;oy = boardy(SQUAREDIM - CHECKERDIM) / 21;// 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 = oxCHECKERDIM / 2;int coy = oyCHECKERDIM / 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 (tmpoxboardxtmpoyboardytmpoxCHECKERDIMboardxBOARDDIMtmpoyCHECKERDIMboardyBOARDDIM){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 * SQUAREDIM1, 8 * SQUAREDIM1);// Paint checkerboard.// 绘制棋盘for (int row = 0; row8; row){g.setColor (((row1) != 0) ? darkGreen : Color.white);for (int col = 0; col8; col){g.fillRect (x1col * SQUAREDIM, y1row * 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);}}
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连连看java代码分析，不过你看连连看java代码分析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; cols6; cols) {
for (int rows = 0; rows5; rows) {
diamondsButton[cols][rows] = new JButton(String
.valueOf(grid[cols1][rows1]));
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() * 251);//button上的数字
for (int alike = 1; alike = 2; alike) {
cols = (int) (Math.random() * 61);
rows = (int) (Math.random() * 51);
while (grid[cols][rows] != 0) {//等于0说明这个空格有了button
cols = (int) (Math.random() * 61);
rows = (int) (Math.random() * 51);
}
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() * 61);
rows = (int) (Math.random() * 51);
while (grid[cols][rows] != 0) {
cols = (int) (Math.random() * 61);
rows = (int) (Math.random() * 51);
}
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; i6; i) {
for (int j = 0; j5; j) {
if (grid[i1][j1] == 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 == secondMsgsecondButton != 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 == y1 || y0 == y - 1))
|| ((x0 == x1 || 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 (yj) {//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 (yj) { // 第二个按钮在空按钮左边
for (i = y1; 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 (x0x) {//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] == 0n == x - 1) {
remove();
}
}
}
if (x0x) {//第一按钮在第二按钮上边
for (n = x0; n = x1; n--) {
if (grid[n][j] != 0) {
k = 0;
break;
}
if (grid[n][j] == 0n == x1) {
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; i8; i) {
if (grid[i][y0] == 0) {// 判断和第一个按钮同列的哪个按钮为空
if (xi) {//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 (xi) {//第二个按钮在这个空按钮的上面
for (j = x1; 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 (y0y) {//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] == 0n == y - 1) {
remove();
}
}
}
if (y0y) {//第二个按钮在第一个按钮左边
for (n = y0; n = y1; n--) {
if (grid[i][n] != 0) {
k = 0;
break;
}
if (grid[i][n] == 0n == y1) {
remove();
}
}
}
}
}//--------------------------------for
}//-------------else
}//------------xiao
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;
} // K=2说明通过了第二次验证
}
}
if (y0j) { // 第一按钮同行空按钮在右边
for (i = y01; i = j; i) {
if (grid[x0][i] != 0) {
k = 0;
break;
} else {
k = 2;
}
}
}
}
public void rowPassOne() {
if (x0i) {//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 (x0i) {//第一个按钮在与它同列的那个空格按钮上面
for (j = x01; 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; cols6; cols) {
for (int rows = 0; rows5; rows) {
if (e.getSource() == diamondsButton[cols][rows])
estimateEven(cols1, rows1, diamondsButton[cols][rows]);
}
}
}
public static void main(String[] args) {
LianLianKan llk = new LianLianKan();
llk.randomBuild();
llk.init();
}
}
连连看JAVA源代码加上 。(初始化代码楼主清洗本身选) 对应在这句话 。别离grid[][]数组的行列即可 ， 你只需定义25个不一样的图片;后面; 这句话是用来设置连连看的图的.setIcon(icons[grid[cols1][rows1])]) 。定义： diamondsButton[cols][rows] = new JButton(String .valueOf(grid[cols1][rows1]))： ImageIcon icons[]= new ImageIcon[25]： diamondsButton[cols][rows].valueOf(grid[cols1][rows1]))diamondsButton[cols][rows] = new JButton(String。它只用了数字; 然后把icons数组初始化对应每个图片即可
如何用JAVA 编写一个连连看游戏全程设计刚试了 。。测试通过 。。
importjavax.swing.*;
importjava.awt.*;
importjava.awt.event.*;
publicclass LianLianKan implements ActionListener {
JFrame mainFrame; // 主面板
Container thisContainer;
JPanel centerPanel, southPanel, northPanel; //子面板
JButton diamondsButton[][] = newJButton[6][5];// 游戏按钮数组
JButton exitButton, resetButton, newlyButton;// 退出，重列 ， 重新开始按钮
JLabel fractionLable = newJLabel("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; cols6; cols) {
for (int rows = 0; rows5; rows) {
diamondsButton[cols][rows] = newJButton(String
.valueOf(grid[cols1][rows1]));
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() * 251);//button上的数字
for (int alike = 1; alike = 2; alike){
cols = (int) (Math.random() * 61);
rows = (int) (Math.random() * 51);
while (grid[cols][rows] != 0) {//等于0说明这个空格有了button
cols = (int) (Math.random() * 61);
rows = (int) (Math.random() * 51);
}
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() * 61);
rows = (int) (Math.random() * 51);
while (grid[cols][rows] != 0) {
cols = (int) (Math.random() * 61);
rows = (int) (Math.random() * 51);
}
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; i6; i) {
for (int j = 0; j5; j) {
if (grid[i1][j1] == 0)
diamondsButton[i][j].setVisible(false);
}
}
}
public void estimateEven(int placeX, intplaceY, 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 == y1 || y0 ==y - 1))
|| ((x0 == x1 || 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 (yj) {//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 (yj) { // 第二个按钮在空按钮左边
for (i = y1; 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 (x0x) {//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] == 0n == x -1) {
remove();
}
}
}
if (x0x) {//第一按钮在第二按钮上边
for (n = x0; n = x1; n--) {
if (grid[n][j] != 0) {
k = 0;
break;
}
if (grid[n][j] == 0n == x1) {
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; i8; i) {
if (grid[i][y0] == 0) {// 判断和第一个按钮同列的哪个按钮为空
if (xi) {//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 (xi) {//第二个按钮在这个空按钮的上面
for (j = x1; 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 (y0y) {//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] == 0n == y -1) {
remove();
}
【连连看java代码分析 连连看 java】}
}
if (y0y) {//第二个按钮在第一个按钮左边
for (n = y0; n = y1; n--) {
if (grid[i][n] != 0) {
k = 0;
break;
}
if (grid[i][n] == 0n == y1) {
remove();
}
}
}
}
}//--------------------------------for
}//-------------else
}//------------xiao
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;
} // K=2说明通过了第二次验证
}
}
if (y0j) { // 第一按钮同行空按钮在右边
for (i = y01; i = j; i) {
if (grid[x0][i] != 0) {
k = 0;
break;
} else {
k = 2;
}
}
}
}
public void rowPassOne() {
if (x0i) {//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 (x0i) {//第一个按钮在与它同列的那个空格按钮上面
for (j = x01; 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; cols6; cols) {
for (int rows = 0; rows5; rows) {
if (e.getSource() ==diamondsButton[cols][rows])
estimateEven(cols1, rows1,diamondsButton[cols][rows]);
}
}
}
public static void main(String[] args) {
LianLianKan llk = new LianLianKan();
llk.randomBuild();
llk.init();
}
}
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