[英]How to convert this method to a boolean
我目前正在尝试实施国际象棋游戏。 我对它进行了结构设计,以便为每种样片类型生成可能的移动并将其存储在数组列表中。 我的木板是2D阵列。 我想知道如何写,如果xTo yTo(点的坐标要移动到)是可能的移动,那么可以进行移动,但是它不会让我使用数组list.contains()
,任何建议都值得赞赏! 这是我所拥有的一个例子。 (用户通过终端输入坐标xFrom,yFrom然后xTo yTo)我现在想知道将其转换为布尔值是否更容易? 并摆脱数组列表?
public Board() {
this.boardsize = DEFAULT_SIZE;
board = new char[boardsize][boardsize];
// Clear all playable fields
for (int x = 0; x < boardsize; x++)
for (int y = 0; y < boardsize; y++)
board[x][y] = FREE;
board[0][7] = BLACKROOK;
board[2][7] = BLACKBISHOP;
board[5][7] = BLACKBISHOP;
board[7][7] = BLACKROOK;
board[0][0] = WHITEROOK;
board[2][0] = WHITEBISHOP;
board[5][0] = WHITEBISHOP;
board[7][0] = WHITEROOK;
对于白嘴鸦...
public ArrayList<int[]> possibleMoves = new ArrayList<int[]>();
public ArrayList<int[]> generatePossibleMoves(char[][] gameBoard, int xFrom, int yFrom) {
for (int i = 1; xFrom + i < gameBoard.length; i++) {
if (getPieceColour(gameBoard, xFrom + i, yFrom) != getPieceColour(gameBoard, xFrom, yFrom)) {
if (gameBoard[xFrom + i][yFrom] != FREE) {
int[] move = {xFrom + i, yFrom};
possibleMoves.add(move);
break; //stops iterating here since a rook is not allowed to jump over other pieces
} else
{
int[] move = {xFrom + i, yFrom};
possibleMoves.add(move);
}
}
}
for (int i = 1; xFrom - i < gameBoard.length; i++) {
if (getPieceColour(gameBoard, xFrom - i, yFrom) != getPieceColour(gameBoard, xFrom, yFrom)) {
if (gameBoard[xFrom - i][yFrom] != FREE) {
int[] move = {xFrom - i, yFrom};
possibleMoves.add(move);
break;
}
else
{
int[] move = {xFrom - i, yFrom};
possibleMoves.add(move);
}
}
}
for (int i = 1; yFrom + i < gameBoard.length+1; i++) { //makes sure the place to be moved is on the board
if (getPieceColour(gameBoard, xFrom + i, yFrom) != getPieceColour(gameBoard, xFrom, yFrom)) {
if (gameBoard[xFrom][yFrom+i] != FREE) {
int[] move = {xFrom, yFrom+i};
possibleMoves.add(move);
break;
}
else
{
int[] move = {xFrom, yFrom+i};
possibleMoves.add(move);
}
}
}
for (int i = 1; yFrom- i < gameBoard.length+1; i++)
if (getPieceColour(gameBoard, xFrom, yFrom - 1) != getPieceColour(gameBoard, xFrom, yFrom)) {
if (gameBoard[xFrom][yFrom - 1] != FREE) {
int[] move = {xFrom, yFrom - 1};
possibleMoves.add(move);
break;
} else {
int[] move = {xFrom, yFrom - 1};
possibleMoves.add(move);
}
}
return possibleMoves;
}
public boolean moveLegal(char[][] gameBoard, int xFrom, int yFrom, int xTo, int yTo){
generatePossibleMoves(gameBoard, xFrom,yFrom);
if(possibleMoves.contains(xTo,yTo){
//this is where I'm stuck
}
}
要检查是否可以移动,一种方法是将这对{xTo,yTo}与您通过generatePossibleMoves函数计算出的所有合法移动进行比较:
public boolean moveLegal(char[][] gameBoard, int xFrom, int yFrom, int xTo, int
yTo){
int[] wantedMove = new int[] {xTo, yTo};
ArrayList<int[]> possibleMoves = generatePossibleMoves(gameBoard, xFrom,yFrom);
boolean isMoveLegal = possibleMoves.stream().anyMatch(possibleMove ->
Arrays.equals(wantedMove, possibleMove));
return isMoveLegal;
}
我将创建另一个可正确实现equals方法的类Coordinates
public class Coordinates {
int x = 0;
int y = 0;
public Coordinates(int x, int y) {
super();
this.x = x;
this.y = y;
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + x;
result = prime * result + y;
return result;
}
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Coordinates other = (Coordinates) obj;
if (x != other.x)
return false;
if (y != other.y)
return false;
return true;
}
}
然后将此类用于ArrayList的类型
所以举个例子
public List<Coordinates> possibleMoves = new ArrayList<Coordinates>();
然后函数变成
public boolean moveLegal(char[][] gameBoard, int xFrom, int yFrom, int xTo, int yTo){
generatePossibleMoves(gameBoard, xFrom,yFrom);
Coordinates checkCoordinates = new Coordinates (xTo,yTo);
if(possibleMoves.contains(xTo,yTo){
...
}
}
最主要的原因是你在保存possibleMoves
阵列,但尝试检查不是一个数组。 如文档中所述 , List.contains()
仅接受1个参数。 由于将数组放入List
,因此您可能需要检查数组: if(possibleMoves.contains({xTo,yTo})
但是实际上,它不能正常工作。 您混合了所有棋子的可能动作,因此可以选择皇后并移动到可以到达骑士的任何位置。
主题 :我建议您使用更多的OOP风格的方法:使用更少的原始数组,使用更多的对象来反映片段。 例如
enum Side {
White; Black;
public Side opposite() {
if (this==White) return Black;
else return White;
}
}
// in separate file
class Pawn {
private ChessSquare currentPosition;
private final Side color;
public boolean couldMoveTo(ChessSquare another) {
if (currentPosition.x == another.x) {
return another.y - currentPosition.y == 1; //TODO check for first move in two sruares
} else if (another.hasPiece(this.color.opposite())) {
// TODO allow to take enemy piece in diagonal
}
}
public List<ChessField> possibleMoves() {
List<ChessField> result = new ArrayList<>();
for (currentSquare in ALL_SQUARES) {
if (couldMoveTo(currentSquare)) result.add(currentSquare)
}
return result;
}
我的示例效率低下,可以通过许多方式进行改进。 另外还有许多其他选项可以组织代码结构。 我想它展示了您如何从一件作品的角度检查一件作品是否可以动弹。 同样,您可以在Pawn
类中隐藏许多细节(例如, 传递规则),同时在较高级别上具有清晰的代码。 您可能会发现possibleMoves()
非常小,实际上对所有部件都是通用的。
PS Chess是一款很棒的游戏,希望您在创建游戏的同时学习象棋和Java。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.