Java 在NXN井字游戏中确定赢家
我尝试用play和Haswen方法创建一个Tic-Tac-Toe游戏类Java 在NXN井字游戏中确定赢家,java,multidimensional-array,tic-tac-toe,Java,Multidimensional Array,Tic Tac Toe,我尝试用play和Haswen方法创建一个Tic-Tac-Toe游戏类 public class TicTacToe { private enum State{Blank, X, O} private int grid; private State[][] board = new State[grid][grid]; private int moves; //Default Constructor public TicTacToe(int grid){ board = new S
public class TicTacToe {
private enum State{Blank, X, O}
private int grid;
private State[][] board = new State[grid][grid];
private int moves;
//Default Constructor
public TicTacToe(int grid){
board = new State[grid][grid];
moves = 0;
}
public void play(State s, int m, int n){
if (board[m][n] == State.Blank){
board[m][n] = s;
}
moves++;
}
public boolean hasWon(State[][] board){
//check winner in rows
boolean state = false;
int j = 0;
int i = 0;
while(j <= grid) {
for (i = 0; i <= grid; i++) {
if (board[j][i] == board[j][i + 1])
state = true;
else state = false;
break;
}
if(state == false)
j++;
else return true;
}
//check winner in columns
while(j <= grid) {
for (i = 0; i < grid; i++) {
if (board[i][j] == board[i + 1][j])
state = true;
else state = false;
break;
}
if (state == true)
j++;
else return true;
}
//check winner in top diagonal
while(j <= grid && i <= grid){
if (board[i][j] == board[i+1][j+1])
state = true;
else state = false;
break;
i++;
j++;
return true;
}
//check winner in bottom diagonal
int k = grid;
int l = grid;
while(k >= 0 && l >= 0){
if (board[k][l] == board[k-1][l-1])
state = true;
else state = false;
break;
k--;
l--;
return true;
}
return false;
}
公共类TicTacToe{
私有枚举状态{Blank,X,O}
私有int网格;
私有国家[]董事会=新国家[电网][电网];
私人内部移动;
//默认构造函数
公共交通(国际电网){
董事会=新国家[电网][电网];
移动=0;
}
公共无效播放(状态s、int m、int n){
如果(线路板[m][n]==状态为空){
板[m][n]=s;
}
移动++;
}
公立学校(州[]委员会){
//按行检查获胜者
布尔状态=假;
int j=0;
int i=0;
而(j当用
for (i = 0; i < n; i++) {
if (board[j][i] == board[j][i + 1])
state = true;
}
什么是“不稳定的”确切地说,这是什么意思?在某些情况下,它是给一个赢家,而在其他情况下则不是。你需要排成一行多少人?3?好吧,首先,你只是比较一条对角线。你可以尝试为方法haswown
实现一些单元测试,并在返回错误结果时找到案例。这是一个非常有用的练习。编辑以更改t结果数组。@Clockwork您编辑它的方式不起作用,因为break
命令现在会导致循环在第一次迭代中直接停止。break
只能在检测到不同的值时使用。
for (i = 0; i < n; i++) {
if (board[j][i] == board[j][i + 1]) {
state = true;
} else {
state = false;
break;
}
}
this.grid = grid;