五子棋java程序的判断胜负的方法

Question

我有这样一段程序：
import java.awt.*;
import java.awt.event.*;
import java.util.*;
class FiveChess extends Canvas
{
LinkedList blacklist = new LinkedList();
LinkedList blacklist1 = new LinkedList();
int count1=0;
int c,c1,r1;
void init(){
this.winner();
this.addListener();
this.setSize(500,500);
this.setBackground(Color.yellow);

}
public void paint(Graphics g){
for(int x=0;x<=500;x=x+20){
g.drawLine(x,0,x,500);
}
for(int y=0;y<=500;y=y+20){
g.drawLine(0,y,500,y);
}
for (int i = 0;i < blacklist.size() ; i++)
{
if(i%2 ==0){
g.setColor(Color.black);
}
else{
g.setColor(Color.white);
}
Block b = (Block)blacklist.get(i);
g.fillOval(b.x,b.y,20,20);
}
}
void addListener(){
this.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e){
count1++;
int x = e.getX();
if((x/10)%2 == 0){
x = (x/10)*10 -10;
}
else{
x = (x/10)*10;
}
c1=x/10;
int y = e.getY();
if((y/10)%2 == 0){
y = (y/10)*10 - 10;
}
else{
y = (y/10)*10;
}
r1=y/10;
Block b = new Block(x,y);
blacklist.add(b);
repaint();
}

});

}
void winner(){

}
public static void main(String[] a){
Frame f=new Frame();
FiveChess ca=new FiveChess();
f.add(ca);
f.setBackground(Color.gray);
f.setLayout(new FlowLayout());
f.setSize(600,600);
f.setVisible(true);
ca.init();
}
}
可以运行处结果，但中间判断胜负的方法没写，请高手指教

Answer 1

每走一步，调用判断方法
思路为：
首先你应该有一个组的集合，就是横、竖、斜三个方向，每5格一组，这个可以根据你的宽高计算出来。

循环这个集合。如果有同色连接5个，那么为胜

同时将五格中已有4个同色并且剩余一格未操作的，放入一个集合内，若上一步没有结果，那么判断这个集合，如果有这个集合有两条以上，并且未操作格坐标不为同一个，那么同样为胜

这么说明白不？

Answer 2

public boolean isWin(int row, int column) {
boolean isWin = false;
char curChess = turn?'@':'o';
if (this.isHorizontalWin(row, column, curChess)
|| this.isVerticalWin(row, column, curChess)
|| this.isLeftObliqueDownWin(row, column, curChess)
|| this.isLeftObliqueUpWin(row, column, curChess)) {
isWin = true;
}
return isWin;
}

public boolean isHorizontalWin(int row, int column, char curChess) {
boolean isWin = false;
int count = 0;
for (int c = column-1; c>=0; c--) {
if (cells[row][c] == curChess) {
count++;
} else {
break;
}
}
for (int c = column+1; c<FiveChess.CELL; c++) {
if (cells[row][c] == curChess) {
count++;
} else {
break;
}
}
System.out.println("horizontal count :"+ count);
if (count >=4) {
isWin = true;
}
return isWin;
}

public boolean isVerticalWin(int row, int column, char curChess) {
boolean isWin = false;

return isWin;
}

public boolean isLeftObliqueDownWin(int row, int column, char curChess) {
boolean isWin = false;

return isWin;
}

public boolean isLeftObliqueUpWin(int row, int column, char curChess) {
boolean isWin = false;
int count = 0;
for (int r = row-1, c = column-1; r>=0 && c>=0; r--, c--){
if (cells[r][c] == curChess) {
count++;
} else {
break;
}
}

if (count >=4) {
isWin = true;
}
return isWin;
}

Answer 3

写个算法
判断矩阵横竖斜。。。，要颜色一样，没走一步都要判定
做的更好点的话，可以连续4个就判定输