所以，我有一个 4x4 的 2D 数组(它总是这些维度).从数组上的一个位置开始，一些行和列，我想找到它的所有有效邻居.到目前为止，我有一个非常笨拙的实现.
So, I have a 4x4 2D array (it will always be these dimensions). Starting with a location on the array, some row and column, I want to find all of its valid neighbors. So far, I have a really clunky implementation.
//add row if ( !((row + 1) > 3)) { //do stuff } //sub row if ( !((row - 1) < 0)) { //do stuff } //add col if ( !((col + 1) > 3)) { //do stuff } //sub col if ( !((col - 1) < 0)) { //do stuff } ... and so on这太残忍了.当我开始知道元素的位置时，我觉得我不需要检查每个邻居.有什么想法吗?
This is brutal. I feel like I do not need to check every single neighbor when I start by knowing the location of the element. Any ideas?
推荐答案对于任何二维数组 cellValues[][] 的 (x,y) 维度下面的代码都可以用于获取任何单元格 (i,j) 的所有 8 个邻居.默认情况下，代码将返回 0.
For any 2D array cellValues[][] of (x,y) dimensions below code can be used for getting all 8 neighbors for any cell (i,j). Code will return 0 by default.
public static ArrayList<Integer> getNeighbors(int i, int j, int x, int y, int[][] cellValues) { ArrayList<Integer> neighbors = new ArrayList<>(); if(isCabin(i, j, x, y)) { if(isCabin(i + 1, j, x, y)) neighbors.add(cellValues[i+1][j]); if(isCabin(i - 1, j, x, y)) neighbors.add(cellValues[i-1][j]); if(isCabin(i, j + 1, x, y)) neighbors.add(cellValues[i][j+1]); if(isCabin(i, j - 1, x, y)) neighbors.add(cellValues[i][j-1]); if(isCabin(i - 1, j + 1, x, y)) neighbors.add(cellValues[i-1][j+1]); if(isCabin(i + 1, j - 1, x, y)) neighbors.add(cellValues[i+1][j-1]); if(isCabin(i + 1, j + 1, x, y)) neighbors.add(cellValues[i+1][j+1]); if(isCabin(i - 1, j - 1, x, y)) neighbors.add(cellValues[i-1][j-1]); } return neighbors; } public static boolean isCabin(int i, int j, int x, int y) { boolean flag = false; if (i >= 0 && i <= x && j >= 0 && j <= y) { flag = true; } return flag; }