2048小游戏 Java 课程设计报告书(5)

protected void do_label_keyPressed(int option) { 字符 字符相等 表示形式 块之和

int a; // a的引入是为了防止连加的情况出现 String str; String str1; int num; switch (option) { case 1: biaoji = 0; // 标记左边能不能合并 for (int i = 0; i < 4; i++) { a = 5; for (int k = 0; k < 3; k++) { // 让左边每2个能合并的全合并了 for (int j = 1; j < 4; j++) {// 遍历16个方块 str = texts[i][j].getText();// 获取当前方块标签文本字符 str1 = texts[i][j - 1].getText();// 获取当前左1方块标签文本 if (str1.compareTo(\如果左1方块文本为空字符 if (str.compareTo(\ biaoji = 1; // 如果向左能移，就标记一下 texts[i][j - 1].setText(str);// 字符左移 setColor(i, j - 1, str); texts[i][j].setText(\当前方块字符置空 setColor(i, j, \ } else if ((str.compareTo(str1) == 0) && (j != a) && (j != a - 1)) {// 如果当前方块和左1方块0文本 num = Integer.parseInt(str); scores += num; times++; str = String.valueOf(2 * num); // 返回 int参数的字符串 texts[i][j - 1].setText(str);// 左1方块文本字符变为两方 setColor(i, j - 1, str); texts[i][j].setText(\当前方块字符置空 setColor(i, j, \ a = j; biaoji = 1; } } } } p1 = 1; Create(); break;

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case 2: biaoji = 0; for (int i = 0; i < 4; i++) { a = 5; for (int k = 0; k < 3; k++) { for (int j = 2; j >= 0; j--) { str = texts[i][j].getText(); str1 = texts[i][j + 1].getText();

if (str1.compareTo(\ if (str.compareTo(\ biaoji = 1; texts[i][j + 1].setText(str); setColor(i, j + 1, str); texts[i][j].setText(\ setColor(i, j, \ } else if (str.compareTo(str1) == 0 && j != a && j != a + 1) { num = Integer.parseInt(str); scores += num; times++; str = String.valueOf(2 * num); texts[i][j + 1].setText(str); setColor(i, j + 1, str); texts[i][j].setText(\ setColor(i, j, \ a = j; biaoji = 1; } } } } p2 = 1; Create(); break; case 3: biaoji = 0; for (int j = 0; j < 4; j++) { a = 5; for (int k = 0; k < 3; k++) { for (int i = 1; i < 4; i++) { str = texts[i][j].getText(); str1 = texts[i - 1][j].getText();

if (str1.compareTo(\

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if (str.compareTo(\ biaoji = 1; texts[i - 1][j].setText(str); setColor(i - 1, j, str); texts[i][j].setText(\ setColor(i, j, \ } else if (str.compareTo(str1) == 0 && i != a && i != a - 1) { num = Integer.parseInt(str); scores += num; times++; str = String.valueOf(2 * num); texts[i - 1][j].setText(str); setColor(i - 1, j, str); texts[i][j].setText(\ setColor(i, j, \ a = i; biaoji = 1; } } } } p3 = 1; Create(); break; case 4: biaoji = 0; for (int j = 0; j < 4; j++) { a = 5; for (int k = 0; k < 5; k++) { for (int i = 2; i >= 0; i--) { str = texts[i][j].getText(); str1 = texts[i + 1][j].getText();

if (str1.compareTo(\ if (str.compareTo(\ biaoji = 1; texts[i + 1][j].setText(str); setColor(i + 1, j, str); texts[i][j].setText(\ setColor(i, j, \

} else if (str.compareTo(str1) == 0 && i != a && i != a + 1) { num = Integer.parseInt(str); scores += num;

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times++; str = String.valueOf(2 * num); texts[i + 1][j].setText(str); setColor(i + 1, j, str); texts[i][j].setText(\ setColor(i, j, \ a = i; biaoji = 1; } } } } p4 = 1; Create(); break; default: break; } if(maxscores<=scores) { maxscores=scores; Recorder.setMaxscores_2048(maxscores); Recorder.keepRecording(); } currentScores.setText(\ labelMaxScores.setText(\最高得分标签 } // 在随机位置产生一个2号或4号方块的方法 public void Create() { int i, j;//用于记录产生的二维方阵的下标值 boolean r = true; String str; if (times > 0 && biaoji == 1) { while (r) { i = random.nextInt(4); j = random.nextInt(4); str = texts[i][j].getText(); if ((str.compareTo(\ int ima = 2 * (int) (1 + Math.random() * 2);// 随机产生2或4 String imass = String.valueOf(ima); texts[i][j].setText(imass); setColor(i, j, imass); times--; r = false;

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p1 = p2 = p3 = p4 = 0; } } //记录当前的状态 for(int p = 0; p < 4; ++p){ for(int q = 0; q < 4; ++q){ temp1[p][q] = temp2[p][q]; if( \ else temp2[p][q] = Integer.parseInt(texts[p][q].getText()); } } temp1scores = temp2scores; temp2scores = scores; } else if (p1 > 0 && p2 > 0 && p3 > 0 && p4 > 0) {// l1到l4同时被键盘赋值为1说明任何方向键都不能产生新的数字，说明游戏失败 tips.setText(\游 戏 结 束 ！\ currentScores.setText(\ JOptionPane.showMessageDialog(null,\游戏结束!\ System.exit(0); } } public void mousePressed(MouseEvent e) { //鼠标按下 //若鼠标位置在灰色的区域内则记录鼠标左键按下时的坐标 if(e.getX()>=20&&e.getX()<=480&&e.getY()>=180&&e.getY()<=640) { this.x1=e.getX(); this.y1=e.getY(); //调试程序输出鼠标的坐标位置 System.out.println(\ \ } } public void mouseReleased(MouseEvent e) { //鼠标释放 //若鼠标位置在灰色区域内并且按下鼠标的位置不为0，则进行相应的处理 if(e.getX()>=20&&e.getX()<=480&&e.getY()>=180&&e.getY()<=640&&this.x1!=0&&this.y1!=0) { this.x2=e.getX() …… 此处隐藏：506字，全部文档内容请下载后查看。喜欢就下载吧 ……