3 回答

TA貢獻1825條經驗 獲得超4個贊
看起來您的代碼對于水平和垂直情況都是正確的。棘手的部分是對角線情況。
讓我們嘗試一下圖片:
在此處輸入圖片說明
對于綠線,您的起始行位置為0 ... maxRow-4。該列將為0 ... startingRow-
偽代碼:
// top-left to bottom-right - green diagonals
for( rowStart = 0; rowStart < rowMax - 4; rowStart++){
count = 0;
int row, col;
for( row = rowStart, col = 0; row < rowMax && col < colMax; row++, col++ ){
if(gridTable[row][col] == player){
count++;
if(count >= 4) return 1;
}
else {
count = 0;
}
}
}
// top-left to bottom-right - red diagonals
for( colStart = 1; colStart < colMax - 4; rowStart++){
count = 0;
int row, col;
for( row = 0, col = colStart; row < rowMax && col < colMax; row++, col++ ){
if(gridTable[row][col] == player){
count++;
if(count >= 4) return 1;
}
else {
count = 0;
}
}
}
對于對角線的另一種方式(從左下到右上),您可以執行類似的操作。

TA貢獻1993條經驗 獲得超6個贊
由于某種原因,我不喜歡計數器,所以我這樣做(它適用于不同尺寸的電路板)。
public boolean areFourConnected(int player){
// horizontalCheck
for (int j = 0; j<getHeight()-3 ; j++ ){
for (int i = 0; i<getWidth(); i++){
if (this.board[i][j] == player && this.board[i][j+1] == player && this.board[i][j+2] == player && this.board[i][j+3] == player){
return true;
}
}
}
// verticalCheck
for (int i = 0; i<getWidth()-3 ; i++ ){
for (int j = 0; j<this.getHeight(); j++){
if (this.board[i][j] == player && this.board[i+1][j] == player && this.board[i+2][j] == player && this.board[i+3][j] == player){
return true;
}
}
}
// ascendingDiagonalCheck
for (int i=3; i<getWidth(); i++){
for (int j=0; j<getHeight()-3; j++){
if (this.board[i][j] == player && this.board[i-1][j+1] == player && this.board[i-2][j+2] == player && this.board[i-3][j+3] == player)
return true;
}
}
// descendingDiagonalCheck
for (int i=3; i<getWidth(); i++){
for (int j=3; j<getHeight(); j++){
if (this.board[i][j] == player && this.board[i-1][j-1] == player && this.board[i-2][j-2] == player && this.board[i-3][j-3] == player)
return true;
}
}
return false;
}

TA貢獻1780條經驗 獲得超4個贊
因此,瀏覽完您的代碼后,似乎對角線檢查只能在一個方向上獲勝(如果我在最低行和最低列中添加標記,會發生什么情況?)
而是,基本檢查算法始終是相同的過程,而不管您要檢查的方向是什么。
您需要一個起點(x / y)和x / y增量(運動方向)。您可以將其歸納為一個方法...
public boolean didWin(int[][] grid, int check, int row, int col, int rowDelta, int colDelta) {
boolean win = true;
for (int count = 0; count < 4; count++) {
if (row < ROWS && row >= 0 && col < COLUMNS && col >= 0) {
int test = grid[row][col];
if (test != check) {
win = false;
break;
}
}
row += rowDelta;
col += colDelta;
}
return win;
}
基本上,您可以從四個方向檢查,也可以向后檢查
所以,如果我們要使用類似...
int[][] gridTable = new int[ROWS][COLUMNS];
gridTable[ROWS - 1][3] = 1;
gridTable[ROWS - 2][3] = 1;
gridTable[ROWS - 3][3] = 1;
gridTable[ROWS - 4][3] = 1;
System.out.println("Vertical");
System.out.println(didWin(gridTable, 1, ROWS - 4, 3, 1, 0) ? "Win" : "Lose");
System.out.println(didWin(gridTable, 1, ROWS - 1, 3, -1, 0) ? "Win" : "Lose");
System.out.println(didWin(gridTable, 1, 0, 3, 1, 0) ? "Win" : "Lose");
gridTable = new int[ROWS][COLUMNS];
gridTable[3][1] = 1;
gridTable[3][2] = 1;
gridTable[3][3] = 1;
gridTable[3][4] = 1;
System.out.println("");
System.out.println("Horizontal");
System.out.println(didWin(gridTable, 1, 3, 1, 0, 1) ? "Win" : "Lose");
System.out.println(didWin(gridTable, 1, 3, 4, 0, -1) ? "Win" : "Lose");
System.out.println(didWin(gridTable, 1, 3, 0, 0, 1) ? "Win" : "Lose");
gridTable = new int[ROWS][COLUMNS];
gridTable[0][1] = 1;
gridTable[1][2] = 1;
gridTable[2][3] = 1;
gridTable[3][4] = 1;
System.out.println("");
System.out.println("Diag");
System.out.println(didWin(gridTable, 1, 0, 1, 1, 1) ? "Win" : "Lose");
System.out.println(didWin(gridTable, 1, 3, 4, -1, -1) ? "Win" : "Lose");
System.out.println(didWin(gridTable, 1, 1, 2, 1, 1) ? "Win" : "Lose");
哪個輸出...
Vertical
Win
Win
Lose
Horizontal
Win
Win
Lose
Diag
Win
Win
Lose
現在,您可以將其總結為...
public boolean didWin(int[][] grid, int check, int row, int col) {
return didWin(grid, check, row, col, 1, 0) ||
didWin(grid, check, row, col, -1, 0) ||
didWin(grid, check, row, col, 0, 1) ||
didWin(grid, check, row, col, 0, -1) ||
didWin(grid, check, row, col, 1, 1) ||
didWin(grid, check, row, col, -1, -1) ||
didWin(grid, check, row, col, -1, 1) ||
didWin(grid, check, row, col, 1, -1);
}
因此,使用類似...
int[][] gridTable = new int[ROWS][COLUMNS];
gridTable[ROWS - 1][3] = 1;
gridTable[ROWS - 2][3] = 1;
gridTable[ROWS - 3][3] = 1;
gridTable[ROWS - 4][3] = 1;
System.out.println("Vertical");
System.out.println(didWin(gridTable, 1, ROWS - 1, 3) ? "Win" : "Lose");
System.out.println(didWin(gridTable, 1, ROWS - 4, 3) ? "Win" : "Lose");
gridTable = new int[ROWS][COLUMNS];
gridTable[3][1] = 1;
gridTable[3][2] = 1;
gridTable[3][3] = 1;
gridTable[3][4] = 1;
System.out.println("");
System.out.println("Horizontal");
System.out.println(didWin(gridTable, 1, 3, 1) ? "Win" : "Lose");
System.out.println(didWin(gridTable, 1, 3, 4) ? "Win" : "Lose");
gridTable = new int[ROWS][COLUMNS];
gridTable[0][1] = 1;
gridTable[1][2] = 1;
gridTable[2][3] = 1;
gridTable[3][4] = 1;
System.out.println("");
System.out.println("Diag");
System.out.println(didWin(gridTable, 1, 0, 1) ? "Win" : "Lose");
System.out.println(didWin(gridTable, 1, 3, 4) ? "Win" : "Lose");
哪個打印出類似...
Vertical
Win
Win
Horizontal
Win
Win
Diag
Win
Win
我要補充一點,這種方法僅在您連續提供4個芯片的正確開始時才有效。例如didWin(gridTable,1,3,3)將為水平檢查提供false而不是true,因為循環只能檢查一個方向。
這樣做的目的不是提供“完整的,即裝即用的”解決方案,而是提供一個可以開發更廣泛的解決方案的概念(我是說,我討厭人們實際上不得不思考;)。我還基于以下想法設計了解決方案:OP將知道最后一塊放置在哪里,即起點;)
通過didWin稍微修改一下方法,就可以從任何點檢查n逐個n網格...
public boolean didWin(int[][] grid, int check, int row, int col, int rowDelta, int colDelta) {
boolean match = false;
int matches = 0;
while (row < ROWS && row >= 0 && col < COLUMNS && col >= 0) {
int test = grid[row][col];
if (test != check && match) {
break;
} else if (test == check) {
match = true;
matches++;
}
row += rowDelta;
col += colDelta;
}
return matches == 4;
}
所以,我用...
public static final int ROWS = 8;
public static final int COLUMNS = 8;
//...
int[][] gridTable = new int[ROWS][COLUMNS];
gridTable[ROWS - 1][3] = 1;
gridTable[ROWS - 2][3] = 1;
gridTable[ROWS - 3][3] = 1;
gridTable[ROWS - 4][3] = 1;
for (int[] row : gridTable) {
StringJoiner sj = new StringJoiner("|", "|", "|");
for (int col : row) {
sj.add(Integer.toString(col));
}
System.out.println(sj);
}
System.out.println(didWin(gridTable, 1, 3, 3));
并能夠使其正常工作。有時答案不是一個完整的解決方案,而是將某人帶到新地方的想法的種子;)
我將進一步增強功能,包括提供預期的連接件數量,但是我很確定這是我真的不需要演示的增強功能;)
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