这是我尝试使用Java重新创建战舰的尝试。我决定只用一艘飞船测试游戏的简单版本，并在游戏板上给飞船一个具体的位置。我发现我的代码有问题。无论我输入什么坐标，我都最终“命中”了这艘船。
这是我到目前为止编写的所有代码：
import java.util.Scanner;
class GameBoard {
Scanner input = new Scanner(System.in); // scanner object
String[][] board = { // game board
{"_", " 1", " 2", " 3", " 4", " 5", " 6", " 7", " 8", " 9", "10"},
{"A", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
{"B", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
{"C", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
{"D", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
{"E", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
{"F", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
{"G", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
{"H", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
{"I", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
{"J", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"}
};
boolean frigateIsAlive = true; // the ship is still alive
int numOfHitsOnFrigate = 0; // number of hits the player made on the frigate
String [] frigate = {board[1][1], board[1][2]}; // ship
public void createBoard(){ // draws the battleship game board
for (int row = 0; row < board.length; row++) {
for (int col = 0; col < board[row].length; col++) {
System.out.print(board[row][col] + "\t");
} // inner loop
System.out.println();
System.out.println();
System.out.println();
} // outer loop
}
public String getUserGuess() { // takes the users guess
System.out.println("Choose a coordinate on the board to fire at");
int x = input.nextInt();
int y = input.nextInt();
String userGuess = board[x][y];
return userGuess;
}
public void checkResult(String userGuess) { // checks the user's guess
if(userGuess.equalsIgnoreCase(frigate[0])){
System.out.println("hit!");
numOfHitsOnFrigate++;
board[1][1] = " *";
createBoard();
}
else if(userGuess.equalsIgnoreCase(frigate[1])) {
System.out.println("hit!");
numOfHitsOnFrigate++;
board[1][2] = " *";
createBoard();
}
else {
System.out.println("miss!");
}
if (numOfHitsOnFrigate == 2) {
System.out.println("Enemy frigate has been sunk!");
frigateIsAlive = false;
}
}
} // end class
public class Game {
public static void run() {
GameBoard newGame = new GameBoard();
newGame.createBoard();
while(newGame.frigateIsAlive) {
newGame.checkResult(newGame.getUserGuess());
}
}
}
public class App {
public static void main(String[] args) {
Game.run();
}
}
最佳答案
由于frigate的声明为：
frigate = {board[1][1], board[1][2]}
，最终将字符串
'[ ]'分配给护卫舰的两个值。然后，在寻找护卫舰并比较值时，会将其与更多空字符串进行比较。可以通过在
[1,2,3,4, n]中将位置x放置在[A,B,C...,Letter_n]中将位置y放置在板上来解决此问题。也就是说，护卫舰的坐标为Frigate.x = 1和Frigate.y = A。我希望这有帮助！
我看到了您进一步实施此问题的问题。我将使Frigate成为具有坐标列表的类：
this.x作为字母或数字的一点this.y作为一点不像您的示例那样键入this.x元组
(this.x, this.y)在列表中会很好对护卫舰列表中的任何其他点执行相同的操作。
在完成护卫舰名单后，还需要更改两件事。
已更改的第一件事是如何检查用户是否在所需范围内调用事物。
必须更改的第二件事是如何确保不会一遍又一遍地调用同一点来“炸毁”一艘船。也就是说，当在护卫舰中的某个点被调用时，应将其从护卫舰中移除。护卫舰中剩余的元组将成为护卫舰上留下的“生命值”的“健康”。回想护卫舰的原始大小，添加
Frigate.initialSize()会非常方便，但这可能会在以后出现。关于java - 战舰代码不起作用，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/26566803/