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20201640
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2013-11-25T19:33:11.187
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2013-11-25T19:33:11.187
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I'm going to assume you have your board as a <code>char[][]</code> or <code>int[][]</code> (matrix), where you might have: <code>if (char[][]) -> 'B' for black, 'R' for red</code> or <code>if (int[][]) -> 1 for black, 0 for red</code> In my mind it would only make sense to have a 2-d array (matrix) for this kind of problem. Anyway, the algorithm to check the winner not just should, but must loop over the board, as others have said. The reason for this is because it is the most elegant solution to this type of problem. You should basically do the following: A nested for loop: one to iterate rows, one to iterate columns. <pre><code>for (int i = 0; i < matrix.length; i ++) { for (int j = 0; j < matrix.length; j ++) { // Check for stuff in here } } </code></pre> You can check vertical, horizontal and diagonal through something like the following: vertical (down): <pre><code>if (colorOfPieceFound) { // check j-1 (move down one row, same column); // check j-2, etc. } </code></pre> horizontal (left): <pre><code>if (colorOfPieceFound) { // check i-1 (move left one column, same row); // check i-2, etc. } </code></pre> diagonal (up-left): <pre><code>if (colorOfPieceFound) { // check [i-1][j+1] (move down one row, same column); // repeat with +/- 2 } </code></pre> Basically you have 8 directions you need to check when you find a piece. You do this for each element of the matrix (i.e. <code>checkAllDirections(matrix[i][j])</code>) or starting at the place where the piece was 'dropped'.
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1. POConnect 4, check for winner algorithm
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1. USChris Cirefice
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1. COYeah thats what I have, a 2d array containing a state which is red or black. Just thought there might be a better way to go about it. Thanks
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 USPhilipGough
2. COWell, I mean since you're using Java you could build your own matrix *iterator*, which would kind of 'functionalize' that process. But I've done similar "winner checking" algorithms for different games, and if the board is anything like a matrix, it's best just to do a nested `for` loop. And like I said in my answer, and as others have said, you could make it '*smarter*' by having it only check for a winner after a move was made, at that particular location (this would make the most sense). Otherwise you could just check the entire board every round. Either one works!
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