我有点卡住试图为我的棋盘制作一个边框，从上到下运行8-1，从左到右运行a-h。我不太清楚该怎么做。建议和帮助非常感谢！干杯：）
目前的输出是：
BR BKn BB BK BQ BB BKn BR
BP BP BP BP BP BP BP BP
以下是Java代码：
public class ex5 {
public enum Chessmen {
WHITE_KING,
WHITE_QUEEN,
WHITE_ROOK,
WHITE_BISHOP,
WHITE_KNIGHT,
WHITE_PAWN,
BLACK_KING,
BLACK_QUEEN,
BLACK_ROOK,
BLACK_BISHOP,
BLACK_KNIGHT,
BLACK_PAWN,
EMPTY
}
public static void printBoard (Chessmen [][] chessboard){
for (int i=0; i<chessboard.length;i++){
for (int j = 0; j<chessboard.length;j++){
switch (chessboard [i][j]){
case WHITE_KING:
System.out.print("WK\t");
break;
case WHITE_QUEEN:
System.out.print("WQ\t");
break;
case WHITE_ROOK:
System.out.print("WR\t");
break;
case WHITE_BISHOP:
System.out.print("WB\t");
break;
case WHITE_KNIGHT:
System.out.print("WKn\t");
break;
case WHITE_PAWN:
System.out.print("WP\t");
break;
case BLACK_KING:
System.out.print("BK\t");
break;
case BLACK_QUEEN:
System.out.print("BQ\t");
break;
case BLACK_ROOK:
System.out.print("BR\t");
break;
case BLACK_BISHOP:
System.out.print("BB\t");
break;
case BLACK_KNIGHT:
System.out.print("BKn\t");
break;
case BLACK_PAWN:
System.out.print("BP\t");
break;
default:
System.out.print(" " + "\t");
break;
}
}
System.out.println("");
}
}
public static void main(String[] args) {
int Rows = 8;
int Columns = 8;
Chessmen [][] chessboard = new Chessmen [Rows][Columns];
for (int i=0;i<chessboard.length; i++){
for (int j=0;j<chessboard[0].length;j++){
if (i==1){
chessboard [i][j]= Chessmen.BLACK_PAWN;
}else if (i==6){
chessboard [i][j]= Chessmen.WHITE_PAWN;
}else if ((i==0&&j==7)||(i==0&&j==0)){
chessboard [i][j]= Chessmen.BLACK_ROOK;
}else if ((i==0&&j==1)||(i==0&&j==6)){
chessboard [i][j] = Chessmen.BLACK_KNIGHT;
}else if ((i==0&&j==2)||(i==0&&j==5)){
chessboard [i][j] = Chessmen.BLACK_BISHOP;
}else if (i==0&&j==3){
chessboard [i][j] = Chessmen.BLACK_KING;
}else if (i==0&&j==4){
chessboard [i][j] = Chessmen.BLACK_QUEEN;
}else if ((i==7&&j==0)||(i==7&&j==7)){
chessboard [i][j]= Chessmen.WHITE_ROOK;
}else if ((i==7&&j==1)||(i==7&&j==6)){
chessboard [i][j] = Chessmen.WHITE_KNIGHT;
}else if ((i==7&&j==2)||(i==7&&j==5)){
chessboard [i][j] = Chessmen.WHITE_BISHOP;
}else if (i==7&&j==3){
chessboard [i][j] = Chessmen.WHITE_KING;
}else if (i==7&&j==4){
chessboard [i][j] = Chessmen.WHITE_QUEEN;
}else {
chessboard [i][j]= Chessmen.EMPTY;
}
}
}
printBoard (chessboard);
}
}
答案 0 :(得分：2)
在外部渲染循环之前和之后调用System.out.println("+--------------------------------+")。
在内部渲染循环之前和之后调用System.out.print("|")。
你需要调整一下。
答案 1 :(得分：2)
正如其他人所说，你需要在一条线上显示空行和空方块。此外，你需要为每个方块打印相同数量的字符，所以我建议你使用BN进行黑夜，将BK用于黑人国王。 N用来表示代数国际象棋符号中的骑士，以区别于国王。
由于我已经在C ++中解决了这个问题，我将在下面发布我的代码。您可以将其用作算法来在Java中进行翻译。它将显示如下板：
(Coups reversibles : 0)
+-X-+---+---+---+---+---+---+-X-+
8 |=R=|=N=|=B=|=Q=|=K=|=B=|=N=|=R=|
+---+---+---+---+---+---+---+---+
7 |=P=|=P=|=P=|=P=|=P=|=P=|=P=|=P=|
+---+---+---+---+---+---+---+---+
6 | | . | | . | | . | | . |
+---+---+---+---+---+---+---+---+
5 | . | | . | | . | | . | |
+---+---+---+---+---+---+---+---+
4 | | . | | . | | . | | . |
+---+---+---+---+---+---+---+---+
3 | . | | . | | . | | . | |
+---+---+---+---+---+---+---+---+
2 | P | P | P | P | P | P | P | P |
+---+---+---+---+---+---+---+---+
1 | R | N | B | Q | K | B | N | R |
=>+-X-+---+---+---+---+---+---+-X-+
a b c d e f g h
它有，我认为一些interesings功能。没有棋子的黑色正方形在它们的中心有一个圆点。电路板底部或顶部的箭头告诉用户下一步要移动哪一侧。一个'X'或没有它告诉你车是否可以城堡。列后面的箭头'^'表示所述列中的pawn可以被称为“en passant”。
所以这里是代码（可能有错误，我只是翻译了法语whitout的标识符和注释再次编译）：
///////////////////////////////////////////////////////////////////////////
std::ostream& operator << (std::ostream& out, const CPlanche& planche)
{
static const string piecesNames[] = {" "," ",
" P ","=P=",
" N ","=N=",
" K ","=K=",
" "," ",
" B ","=B=",
" R ","=R=",
" Q ","=Q="};
// We display how many revirsable moves have been playes
out <<" (Coups reversibles : " <<board.RecersiblesMoves() <<")\n";
// If it's black to move we display an arrow
if (board.ColorToMove() == BLACK)
{
out <<"=>";
}
else
{
out <<" ";
}
// We display the top line
out <<"+-" <<(Board.Castling(Black, QueenSideCastling)?'X':'-') <<"-+---+---+---+---+---+---+-" <<(board.Castling(Black, LingSideCastling)?'X':'-') <<"-+\n";
// We display all 8 lines
for (int line = 0; line < 8; line++)
{
out <<(char)('8' - line) <<' ';
// for each column
for (int column = 0; column < 8; column++)
{
out <<'|';
if (board[56 - 8 * line + column] != EMPTY)
{
out <<piecesNames[board[56 - 8 * iLigne + column]];
}
else
{
// If both the line and column are even OR if both are odds
if (!((line | column) & 1) || (line & column & 1))
{
out <<" ";
}
else
{
out <<" . ";
}
}
}
out <<"|\n";
if (line != 7)
{
out <<" +---+---+---+---+---+---+---+---+\n";
}
}
// If it's white to move we display an arrow
if (board.ColorToMove() == WHITE)
{
out <<"=>";
}
else
{
out <<" ";
}
// We display the bottom line
out <<"+-" <<(planche.Castling(WHITE, CastlingQueenSide)?'X':'-') <<"-+---+---+---+---+---+---+-" <<(Board.Castling(WHITE, CastlingKingSide)?'X':'-') <<"-+\n";
// Whe display the arrow for the prise en passant if there is one.
if (board.ColumnPriseEnPassant() != NO_PRISE_EN_PASSANT)
{
for (int i = 0; i < (board.ColumnPriseEnPassant() + 1); i++)
{
out <<" ";
}
out <<"^\n";
}
// We display the columns letters
out <<" a b c d e f g h\n";
return out;
}
我希望这会有所帮助。
熔点
答案 2 :(得分：0)
[编辑] 回应你的评论...
我认为你想要的实际上是一个10x10网格，你需要调整你的逻辑，只在10x10内打印游戏区域为8x8，并使用外部方块（不包括角落）打印你的字母和数字。 [结束编辑]
你的默认值不是EMPTY它只是什么都没有。在您的交换机中捕获EMPTY案例并打印您的标签，您将看到所有空白区域
换句话说，而不是case BLACK_PAWN:
System.out.print("BP\t");
break;
default:
System.out.print(" " + "\t");
break;
}
你想做的事：
case BLACK_PAWN:
System.out.print("BP\t");
break;
case EMPTY:
System.out.print(" " + "\t");
break;
}
答案 3 :(得分：0)
我为Java中的N皇后问题做了这个。所以，如果你试图打印除了女王之外的其他作品，那么相应地定制你的代码。如果您不熟悉这个问题，那么将N个皇后放置在N×N棋盘上是一项挑战，这样就不会有两个皇后互相检查。我知道这是你发布问题后的几年，但我想我会把代码片段放在这里供将来使用的人。
无论如何，第一步是创建我们的N x N数组，表示棋盘上的空格以及当前在棋盘上的棋子的价值。值得注意的是，由于这是Java，我们的电路板是基于0的（因为每行中的空格编号为0到7，而不是1到8）：
public static final int N = 8;
public static char [][] board = new char[N][N];
//initialize every space to a space
for(int i = 0; i < N; i++){
for(int j = 0; j < N; j++){
board[i][j] = ' ';
}
}
看到我们将每个点初始化为一个空格，这样我们就可以迭代我对于皇后位置的向量，并且只改变女王需要放置的位置。
以下是解决方案的示例向量：[1 6 2 5 7 4 0 3]。假设这个向量由int[] vector = new int[N]表示。元素的数量等于板中的行数。它的向量中的索引对应于行，并且值本身对应于后者中行后的列号。所以我们只需循环遍历这些并相应地更新电路板阵列：
for(int j = 0; j < N; j++){
board[j][vector[j]] = 'Q';
}
如果您的问题涉及实际的国际象棋游戏，您需要遍历每一个棋子并以类似的方式更新其棋盘位置。
在我解释如何制作这个电路板之前，让我向您展示与前面提到的矢量相对应的电路板。另外，请注意没有皇后检查！我用遗传算法解决了这个问题。
+---+---+---+---+---+---+---+---+
| | Q | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | Q | |
+---+---+---+---+---+---+---+---+
| | | Q | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | Q | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | Q |
+---+---+---+---+---+---+---+---+
| | | | | Q | | | |
+---+---+---+---+---+---+---+---+
| Q | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | Q | | | | |
+---+---+---+---+---+---+---+---+
我们的下一步是在行之间（以及在顶部和底部）创建边框。我将每一行视为格式化字符串。由于我们有空格行和边框行，因此我们的行数组大小为2 * N + 1。
String[] row = new String[2*N + 1];
String border = "";
//must be the length of board
for(int i = 0; i < N; i++){
border += "+---";
}
border += "+";
//every other row is a border row
for(int i = 0; i < 2*N +1; i+=2){
row[i] = border;
}
//must include the bottom
row[2*N] = border;
接下来，我们为电路板空间自己创建字符串。由于我们已经在我们的板[] []中有这些，我们只需要格式化。
for(int i = 0; i < N; i++){
//left bar
String spaces = "| ";
//place enclosing right bar and spaces so next char goes in middle of next space
for(int j = 0; j < N; j++){
spaces += board[i][j];
spaces += " | ";
}
//add the spaces string to the rows at the odd indices
row[2*k +1 ] = spaces;
}
剩下的就是打印那块板子。
for(int i = 0; i < 2*N +1; i++){
System.out.println(row[i]);
}
希望这有助于某人！