java基于swing实现的连连看代码

1、java基于swing实现的连连看代码这篇文章主要介绍了java基于swing实现的连连看代码,包含了游戏中涉及的事件处理与逻辑功能,需要的朋友可以参考下本文实例讲述了java基于swing实现连连看代码。分享给大家供大家参考。 主要功能代码如下： 复制代码 代码如下: package llkan; import javax.swing.*; import java.awt.*; import java.awt.event.*; /* * 连连看游戏 * author administrator *2014年10月17日 */ public class maingame implements

2、actionlistener jframe mainframe; / 主面板 container thiscontainer; jpanel centerpanel, southpanel, northpanel; / 子面板 jbutton diamondsbutton = new jbutton65;/ 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, secondbutto

3、n; / 分别记录两次被选中的按钮 int grid = new int87;/ 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(连连看游戏); thiscontainer = mainframe.getconten

4、tpane(); 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 (

5、int cols = 0; cols 6; cols+) for (int rows = 0; rows 5; rows+) diamondsbuttoncolsrows = new jbutton( string.valueof(gridcols + 1rows + 1); diamondsbuttoncolsrows.addactionlistener(this); centerpanel.add(diamondsbuttoncolsrows); exitbutton = new jbutton(退出); exitbutton.addactionlistener(this); resetb

6、utton = 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(); no

7、rthpanel.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() *

8、6 + 1); rows = (int) (math.random() * 5 + 1); while (gridcolsrows != 0) cols = (int) (math.random() * 6 + 1); rows = (int) (math.random() * 5 + 1); this.gridcolsrows = randoms; public void fraction() fractionlable.settext(string.valueof(integer.parseint(fractionlable .gettext() + 100); public void r

9、eload() int save = new int30; int n = 0, cols, rows; int grid = new int87; for (int i = 0; i = 6; i+) for (int j = 0; j = 0) cols = (int) (math.random() * 6 + 1); rows = (int) (math.random() * 5 + 1); while (gridcolsrows != 0) cols = (int) (math.random() * 6 + 1); rows = (int) (math.random() * 5 + 1

10、); this.gridcolsrows = saven; 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 (gridi + 1j + 1 = 0) diamondsbuttonij.setvisible(false); public void estimateeven(int placex, int placey, jbutton bz) if (pressinfo

11、rmation = false) x = placex; y = placey; secondmsg = gridxy; secondbutton = bz; pressinformation = true; else x0 = x; y0 = y; fristmsg = secondmsg; firstbutton = button; x = placex; y = placey; secondmsg = gridxy; secondbutton = bz; if (fristmsg = secondmsg & secondbutton != firstbutton) xiao(); pub

12、lic 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 j) / 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 (gridxi != 0) k = 0; break; else k = 1; / k=1说明通过

13、了第一次验证 if (k = 1) linepassone(); if (y j) / 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 (gridxi != 0) k = 0; break; else k = 1; if (k = 1) linepassone(); if (y = j) linepassone(); if (k = 2) if (x0 = x) remove(); if (x0 x) for (n = x0; n x) for (n = x0; n

14、= x + 1; n-) if (gridnj != 0) k = 0; break; if (gridnj = 0 & n = x + 1) remove(); for (i = 0; i i) for (j = x - 1; j = i; j-) if (gridjy != 0) k = 0; break; else k = 1; if (k = 1) rowpassone(); if (x i) for (j = x + 1; j = i; j+) if (gridjy != 0) k = 0; break; else k = 1; if (k = 1) rowpassone(); if

15、 (x = i) rowpassone(); if (k = 2) if (y0 = y) remove(); if (y0 y) for (n = y0; n y) for (n = y0; n = y + 1; n-) if (gridin != 0) k = 0; break; if (gridin = 0 & n = y + 1) remove(); 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 (gridx0i != 0) k

16、 = 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与第二按钮之间 for (i = y0 + 1; i i) for (j = x0 - 1; j = i; j-) if (gridjy0 != 0) k = 0; break; else k = 2; if (x0 i) for (j = x0 + 1; j = i; j+) if (gridjy0 != 0) k = 0; break; else k = 2; public void remove() firstbutton.setvisible(false); s

17、econdbutton.setvisible(false); fraction(); pressinformation = false; k = 0; gridx0y0 = 0; gridxy = 0; public void actionperformed(actionevent e) if (e.getsource() = newlybutton) int grid = new int87; this.grid = grid; randombuild(); mainframe.setvisible(false); pressinformation = false; init(); if (e.getsource() = exitbutton) system.exit(0); if (e.getsource() = resetbutton) relo