我有一个问题，我一直想弄清楚，但我被卡住了。基本上我一直试图在不是国际象棋的游戏中实现车辆运动的逻辑，但我坚持下去。我会告诉你详情：
我的车现在做的事情是，它可以向一个方向但它可以在那条线上的任何地方。我需要帮助，试图找出如何添加更多逻辑，以确保它只能在路径清晰的情况下进行。这是一个例子：
小“p”和“r”是玩家2的棋子，大“P”和“R”是玩家的棋子。现在，右上角的R（车）只能向右移动，但是如果你这样做，它将超越当兵，然后可以尽可能地向下移动。
* R R R *
* P P P *
* * * * *
* p p p *
* r r r *
以下是我对车辆的代码：
public boolean isLegalMove(Location from,Location to)
{
// Row is XPosition (Up/Down)
// Column is YPosition(Left/Right)
int fromRow = from.getXPosition();
int fromColumn = from.getYPosition();
int toRow = to.getXPosition();
int toColumn = to.getYPosition();
// higher row or column or both
if(((fromColumn >= toColumn) || (fromColumn <= toColumn)) && ((fromRow == toRow))) {
return true;
}
if(((fromRow >= toRow) || (fromRow <= toRow)) && ((fromColumn == toColumn))) {
return true;
}
return false;
}
我想如果路径中有任何东西，我会打算另一种检查逻辑的方法，称之为isPathClear()
编辑：
这是代码的其余部分：
public class Board
{
// The depth and width of the field.
public static final int ROW = 5;
public static final int COLUMN = 5;
public static final String EMPTYPIECE = " * ";
//Storage for the game pieces
private GamePiece [] [] gameBoard;
//Makes the balls and torches for player1
private Pawn1 p1Pawn1,p1Pawn2,p1Pawn3;
private Rook1 p1Rook1,p1Rook2,p1Rook3;
//Makes the ball and torchers for player2
private Pawn2 p2Pawn1,p2Pawn2,p2Pawn3;
private Rook2 p2Rook1,p2Rook2,p1Rook3;
/**
* Makes a 5x5 Gameboard
*/
public Board()
{
// initialise instance variables
gameBoard = new GamePiece [ROW][COLUMN];
//Makes pieces for player1
p1Pawn1 = new Pawn1();
p1Pawn2 = new Pawn1();
p1Pawn3 = new Pawn1();
p1Rook1 = new Rook1();
p1Rook2 = new Rook1();
p1Rook3 = new Rook1();
//Makes pieces for player2
p2Pawn1 = new Pawn2();
p2Pawn2 = new Pawn2();
p2Pawn3 = new Pawn2();
p2Rook1 = new Rook2();
p2Rook2 = new Rook2();
p2Rook3 = new Rook2();
}
/**
* Makes new games
*/
public void newGame()
{
// Assigns the piece of the board for player1
gameBoard[0][1] = p1Rook1;
gameBoard[0][2] = p1Rook2;
gameBoard[0][3] = p1Rook3;
gameBoard[1][1] = p1Pawn1;
gameBoard[1][2] = p1Pawn2;
gameBoard[1][3] = p1Pawn3;
// Assigns the pieces of the board for player2
gameBoard[4][1] = p2Rook1;
gameBoard[4][2] = p2Rook2;
gameBoard[4][3] = p2Rook3;
gameBoard[3][1] = p2Pawn1;
gameBoard[3][2] = p2Pawn2;
gameBoard[3][3] = p2Pawn3;
}
/**
* Displays the content of the board
*/
public void displayBoard()
{
System.out.println(" a b c d e");
int counter = 1;
for (int i = 0; i < gameBoard.length; i++){
System.out.print(counter);
for (int j = 0; j < gameBoard[i].length; j++) {
if (gameBoard[i][j] == null) {
System.out.print(EMPTYPIECE);
} else {
System.out.print(" " + gameBoard[i][j] + " ");
}
}
counter++;
System.out.println();
}
}
/**
* Moves the movepiece from one locatin to another
* @param from - where the location was from
* @param to - Where the location is going to
*/
public void movePiece(Location from,Location to) throws InvalidMoveException
{
int fromRow = from.getXPosition();
int fromColumn = from.getYPosition();
int toRow = to.getXPosition();
int toColumn = to.getYPosition();
if (gameBoard[fromRow][fromColumn] == null) {
throw new InvalidMoveException("Invalid input for source location.");
}
if (! checkBounds(from, to)) {
throw new InvalidMoveException("Invalid input for destination location.");
}
if (isSameLocation(from, to)){
throw new InvalidMoveException("Invalid move, source and destination cannot bethe same.");
}
if (! gameBoard[fromRow][fromColumn].isLegalMove(from, to)) {
throw new InvalidMoveException("Invalid move for this piece.");
}
gameBoard[toRow][toColumn] = gameBoard[fromRow][fromColumn];
gameBoard[fromRow][fromColumn] = null;
displayBoard();
}
/**
* Checks a proposed move to ensure it is within the bounds of the board.
* @param source location, destination location
* @return true if both source and destination are within bounds
*/
private boolean checkBounds(Location from, Location to)
{
int fromRow = from.getXPosition();
int fromColumn = from.getYPosition();
int toRow = to.getXPosition();
int toColumn = to.getYPosition();
boolean testFrom = (fromRow >= 0) && (fromColumn >= 0) && (fromRow < gameBoard.length) && (fromColumn < gameBoard[0].length);
boolean testTo = (toRow >= 0) && (toColumn >= 0) && (toRow < gameBoard.length) && (toColumn < gameBoard[0].length);
return testFrom && testTo;
}
/**
* Checks a proposed move to ensure source and destination are different.
* @param source location, destination location
* @return true if source and destination are the same
*/
private boolean isSameLocation(Location from, Location to)
{
int fromRow = from.getXPosition();
int fromColumn = from.getYPosition();
int toRow = to.getXPosition();
int toColumn = to.getYPosition();
return fromRow == toRow && fromColumn == toColumn;
}
答案 0 :(得分：4)
如果不知道主板上还有什么，你就无法知道路径是否清晰。但是，您的方法签名不会使此功能访问板的布局。如果你通过整个棋盘，你可以使用一个循环检查其间的所有方块。
来自Lord Torgamus：
你不会检查董事会是否为空。您必须检查车的源位置和目的地位置之间的各个空间。
现在我知道董事会的样子，这里有一些代码：
public boolean isLegalMove(Location from,Location to)
{
// Row is XPosition (Up/Down)
// Column is YPosition(Left/Right)
int fromRow = from.getXPosition();
int fromColumn = from.getYPosition();
int toRow = to.getXPosition();
int toColumn = to.getYPosition();
// Has to be same row or column
if(fromRow != toRow || fromColumn != toColumn) return false;
// Can't move to the same square
if(fromRow == toRow && fromColumn == toColumn) return false;
// Rows are the same
if(fromRow - toRow == 0) {
// this will hold the column of the we're going to check next
int newPos = fromColumn;
// Should we go up or down?
int amount = (toColumn - fromColumn < 0) ? -1 : 1;
while(newPos != toColumn) {
newPos += amount;
// if it's not null, we found a different piece
if(gameBoard[fromRow][newPos] != null) return false;
}
if(gameBoard[toRow][toColumn] != null) {
// return false if it's your own piece, true if it's not
}
// Columns are the same
} else {
// this will hold the row of the we're going to check next
int newPos = fromRow;
// Should we go up or down?
int amount = (toRow - fromRow < 0) ? -1 : 1;
while(newPos != toRow) {
newPos += amount;
// if it's not null, we found a different piece
if(gameBoard[newPos][fromColumn] != null) return false;
}
if(gameBoard[toRow][toColumn] != null) {
// return false if it's your own piece, true if it's not
}
}
return true;
}
编辑你希望能够捕获对手的棋子的情况......但我没有把最后一点代码放入，因为你必须再次更改方法签名。寻找我的评论。另请注意，它现在是while循环，而不是do-while。