あなたはかなり良いスタートをしていますが、私は計画していたよりも多くの変更を加えました。
ご質問がある場合は、セルと呼ばれる単純なOOPクラスを使用して値とセルの位置を追跡し、表示機能を提供し、多くの変数とハードコードされた数値を変換しました定数に（最終的に始まり、プログラム全体で同じ値を持つ）
私はあなたに勝利条件を残していることがわかります！
私はこのことが有用であると思っています。慣れ親しんでいるコンテキストで使用されている2D配列が理解できるように思えます。処理のための2Dアレイを使用
マイ通常のプロセス： 使用セットアップ（）（初期アレイは 使用ドロー値に設定する）が表示機能、配列データを変更するために他の機能を使用するアイテムの表示機能 の各々を呼び出しますセルは
メインファイルを表示する方法を知っているの
// Grid Size constants
final int GRID_COLUMNS = 10;
final int GRID_ROWS = 8;
// Display constants
final int CELL_SCALE = 80;
final int COIN_RADIUS = 75;
final int BOARD_PADDING = 100;
final color[] PLAYER_COLORS = new color[]{color(0), color(200, 10, 10), color(200, 200, 10)};
// cell values
final int EMPTY_CELL = 0;
final int PLAYER_1 = 1;
final int PLAYER_2 = 2;
// game data
Cell[][] grid = new Cell [GRID_COLUMNS][GRID_ROWS];
int currentPlayer;
void setup() {
size(1000, 800);
ellipseMode(CENTER); // only needs to be set once per sketch
setupBlankBoard();
}
// method to populate the array with blank cells, used to initiate and reset the game
void setupBlankBoard() {
currentPlayer = PLAYER_1; // reset game and start with first player
// populate array
for (int y=0; y < GRID_ROWS; y++) { // for every vertical row,
for (int x=0; x < GRID_COLUMNS; x++) { // for every horizontal cell in that row
// rather than translates I prefer to calculate the actual x,y position and just display it there,
// we add the padding offset to every cell, and then take the column/row times the width of the square cell
grid[x][y] = new Cell(BOARD_PADDING+x*CELL_SCALE, BOARD_PADDING+y*CELL_SCALE);
}
}
}
void changePlayers() {
if (currentPlayer == PLAYER_1) {
currentPlayer = PLAYER_2;
} else {
// already was player 2, so change to 1
currentPlayer = PLAYER_1;
}
}
boolean placeCoin(int column) {
boolean coinPlaced = false;
// now we know the column, we need to find the lowest cell in that column that is empty
for (int y = GRID_ROWS-1; y >= 0; y--) { // let's start at bottom and move up to reduce computations
// for every cell, test if it is empty
if (grid[column][y].isEmpty()) {
// if found a valid cell, place current players token and exit the loop (break)
grid[column][y].setValue(currentPlayer);
coinPlaced = true;
break;
}
}
return coinPlaced;
}
void checkCoins() {
// scan the array for 4 of the same value in a row
for (int y=0; y < GRID_ROWS; y++) { // for every vertical row,
for (int x=0; x < GRID_COLUMNS; x++) { // for every horizontal cell in that row
//grid[x][y]
// I will leave this to you ;)
// keep in mind to check neighbors all you need to do is add or subtract 1 from the x or y value
// however when you are at the edge of the board be careful not to try and look at a neighbor that is off the edge, you will get a null pointer or an array out of bounds exception
if (x+1<GRID_COLUMNS) {
Cell toTheRight = grid[x+1][y];
}
// for each cell you could look at the 3 above the current, 3 below the current, 3 to the right and 3 to the left, 3 diagnal in each direction and then manually try and find 4 adjacent same color cells
// or a bit more complicated is a recursive solution that checks its 8 immediate neighbor and for each that match the center cell run the same function to test its 8 neighbors keeping track of the current inARowCount and returning true when it is found.
// would be a bit hard because you would need to make sure the second cell doesn't follow back to the original cell, and start an endless loop
}
}
}
void draw() {
background(1, 1, 1);
noStroke();
// draw all cells
for (int y=0; y < GRID_ROWS; y++) { // for every vertical row,
for (int x=0; x < GRID_COLUMNS; x++) { // for every horizontal cell in that row
grid[x][y].display(); // draw this cell
}
}
// draw next coin floating above the board, contrain its positon to above the board
fill(PLAYER_COLORS[currentPlayer]);
int currentMouseX = constrain(mouseX, BOARD_PADDING+COIN_RADIUS/2, BOARD_PADDING+(GRID_COLUMNS*CELL_SCALE)-COIN_RADIUS/2);
//currentMouseX = 40*(ceil(abs(currentMouseX/40)));
ellipse(currentMouseX, BOARD_PADDING/2, COIN_RADIUS, COIN_RADIUS);
}
// press any key to rest the game, probably want to test for a certain key here!
void keyPressed() {
setupBlankBoard();
}
// on press attempt to place a coin
void mousePressed() {
int currentMouseX = constrain(mouseX, BOARD_PADDING+COIN_RADIUS/2, BOARD_PADDING+(GRID_COLUMNS*CELL_SCALE)-COIN_RADIUS/2);
// determine what column we are over
int column = (currentMouseX - BOARD_PADDING)/CELL_SCALE;
// if choice is a valid coin placement
if (placeCoin(column)) {
// toggle players if a coin was placed
changePlayers();
// after each placement check win conditions
checkCoins();
}
}
セルクラス
class Cell {
int x, y;
int value; // 0 for empty, 1 for team 1, 2 for team 2 (constants defined at top of main file)
Cell(int x, int y) {
// default constructor to create empty cell
this(x, y, EMPTY_CELL);
}
// allows cell value to be set at creation
Cell(int x, int y, int value) {
this.x = x;
this.y = y;
this.value = value;
}
boolean setValue(int value){
value = constrain(value, EMPTY_CELL,PLAYER_2); // keeps it from setting a value outside of our range
if(this.value == EMPTY_CELL){
this.value = value;
return true; // placed
}
return false; // was not able to place it as there is already a value
}
boolean isEmpty(){
return this.value == EMPTY_CELL;
}
void display() {
fill(0, 0, 255);
rect(this.x, this.y, CELL_SCALE, CELL_SCALE);
// Draw Circle color based on current value, could simply just put fill(playerColors[this.value]); but this seems a bit more clear
if(this.value == EMPTY_CELL){
fill(PLAYER_COLORS[EMPTY_CELL]);
}else if(this.value == PLAYER_1){
fill(PLAYER_COLORS[PLAYER_1]); // red
}else if(this.value == PLAYER_2){
fill(PLAYER_COLORS[PLAYER_2]); // yellow
}
ellipse(this.x + CELL_SCALE/2, this.y + CELL_SCALE/2, COIN_RADIUS, COIN_RADIUS);
}
}