1、.【連連看】簡(jiǎn)單小游戲初始源代碼使用說(shuō)明：進(jìn)入java開(kāi)發(fā)工具eclipse 中，在左面建立一個(gè)任意名的java project, 然后在工程下建立一個(gè)名字為lianliankan 的 package，最后建立一個(gè)名字為lianliankan 的 class 類。將以下程序代碼復(fù)制入class 內(nèi)容面板中，點(diǎn)擊運(yùn)行。package lianliankan;import javax.swing.*;import java.awt.*;import java.awt.event.*;public class lianliankan implements actionlistenerjframe m

2、ainframe; / 主面板container thiscontainer;jpanel centerpanel,southpanel,northpanel; / 子面板jbutton diamondsbutton = new jbutton65;/游戲按鈕數(shù)組jbutton exitbutton,resetbutton,newlybutton; /退出，重列，重新開(kāi)始按鈕jlabel fractionlable=new jlabel(0); / 分?jǐn)?shù)標(biāo)簽jbutton firstbutton,secondbutton; /分別記錄兩次被選中的按鈕int grid = new int87;/

3、儲(chǔ)存游戲按鈕位置static boolean pressinformation=false; /判斷是否有按鈕被選中int x0=0,y0=0,x=0,y=0,fristmsg=0,secondmsg=0,validatelv; / 游戲按鈕的位置坐標(biāo)int i,j,k,n;/ 消除方法控制public void init()mainframe=new jframe(jkj連連看 );thiscontainer = mainframe.getcontentpane();thiscontainer.setlayout(new borderlayout();centerpanel=new jpan

4、el();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+ )diamondsbuttoncolsrows=new jbutton

5、(string.valueof(gridcols+1rows+1); diamondsbuttoncolsrows.addactionlistener(this); centerpanel.add(diamondsbuttoncolsrows);exitbutton=new jbutton( 退出 );exitbutton.addactionlistener(this);resetbutton=new jbutton( 重列 );resetbutton.addactionlistener(this);newlybutton=new jbutton( 再來(lái)一局 );.newlybutton.ad

6、dactionlistener(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

7、() 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(gridcolsrows!=0) cols=(int)(math.random()*6+1);rows=(int)(math.random()*5+1);this.gridcolsrows=randoms;public

8、void fraction()fractionlable.settext(string.valueof(integer.parseint(fractionlable.gettext()+100);public void reload() 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) c

9、ols=(int)(math.random()*6+1);rows=(int)(math.random()*5+1);this.gridcolsrows=saven;n-;mainframe.setvisible(false);pressinformation=false; / 這里一定要將按鈕點(diǎn)擊信息歸為初始 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 pl

10、acey ,jbutton bz) if(pressinformation=false) x=placex;y=placey;secondmsg=gridxy;secondbutton=bz;pressinformation=true;else x0=x;y0=y;fristmsg=secondmsg;firstbutton=secondbutton;x=placex;y=placey;secondmsg=gridxy;secondbutton=bz;if(fristmsg=secondmsg & secondbutton!=firstbutton)xiao();public void xia

11、o() / 相同的情況下能不能消去。仔細(xì)分析，不一條條注釋if(x0=x &(y0=y+1|y0=y-1) | (x0=x+1|x0=x-1)&(y0=y) / 判斷是否相鄰 remove();else.for (j=0;jj) / 如果第二個(gè)按鈕的 y 坐標(biāo)大于空按鈕的 y 坐標(biāo)說(shuō)明第一按鈕在第二按鈕左邊 for (i=y-1;i=j;i- ) / 判斷第二按鈕左側(cè)直到第一按鈕中間有沒(méi)有按鈕if (gridxi!=0) k=0;break;else k=1; /k=1 說(shuō)明通過(guò)了第一次驗(yàn)證if (k=1) linepassone();if (yj) / 如果第二個(gè)按鈕的 y 坐標(biāo)小于空按鈕的

12、 y 坐標(biāo)說(shuō)明第一按鈕在第二按鈕右邊 for (i=y+1;i=j ;i+ ) / 判斷第二按鈕左側(cè)直到第一按鈕中間有沒(méi)有按鈕 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 (x0x) for (n=x0;nx) for (n=x0;n=x+1 ;n- ) if (gridnj!=0) k=0;break;if(gridnj=0 & n=x+1) remove();for (i=0;ii) for (j=x-1;j=i

13、;j- ) if (gridjy!=0) k=0;break;else k=1; if (k=1) rowpassone();if (xi) for (j=x+1;j=i;j+ ) if (gridjy!=0) k=0;break;else k=1; if (k=1) rowpassone();if (x=i) rowpassone();.if (k=2)if (y0=y) remove();if (y0y) for (n=y0;ny) for (n=y0;n=y+1 ;n-) if (gridin!=0) k=0;break;if(gridin=0 & n=y+1) remove();pub

14、lic void linepassone()if (y0j) / 第一按鈕同行空按鈕在左邊f(xié)or (i=y0-1;i=j ;i- ) / 判斷第一按鈕同左側(cè)空按鈕之間有沒(méi)按鈕 if (gridx0i!=0) k=0;break;else k=2; /k=2 說(shuō)明通過(guò)了第二次驗(yàn)證if (y0j) / 第一按鈕同行空按鈕在與第二按鈕之間for (i=y0+1;ii) for (j=x0-1;j=i ;j- ) if (gridjy0!=0) k=0;break;else k=2; if (x0i) for (j=x0+1;j=i ;j+ ) if (gridjy0!=0) k=0;break;el

15、se k=2; public void remove()firstbutton.setvisible(false);secondbutton.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)reload();for(int cols