因此，我有一个4x4的二维数组（它始终具有这些维度）。从数组上的某个位置开始，一些行和列，我想找到它的所有有效邻居。到目前为止，我的实现非常笨拙。
//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
这太残酷了。我感觉开始时只需要知道元素位置，就无需检查每个相邻项。有什么想法吗？
对于任何一个二维数组cellValues[][]，其维度为(x,y)，下列代码可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。
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;
}
for (int coff = -1; coff < 3; coff += 2) {
for (int roff = -1; roff < 3; roff += 2) {
if ( col + coff >= 0 &&
col + coff < array.length &&
row + roff >= 0 &&
row + roff < array[row].length) {
// do stuff with array[col + coff][row + roff]
}
}
}
这个循环结构会将列和行的偏移量从-1翻转到1，然后在第三次迭代时停止。
但请注意，在您的代码中，检查!(stuff) > 4会导致ArrayIndexOutOfBounds异常，因为请记住最后一个索引是4-1。
这是我的做法：编写一个方法，该方法获取x，y坐标对的列表以获得有效邻居，给定任意的[x，y]点并推广到任何数组大小：
public List<int[]> getNeighbors(x, y, maxX, maxY) {
neighbors = new ArrayList<int[]>;
if x > 0:
neighbors.add({x-1, y});
if y > 0:
neighbors.add({x, y-1});
if x < maxX:
neighbors.add({x+1, y});
if x < maxY:
neighbors.add({x, y+1});
return neighbors;
}
[...]
for (int[] coords : getNeighbors(x, y, 4, 4)) {
// do stuff
}
// do stuff 的功能是什么，它都是最可读的语法。 - Radiodef什么构成有效的邻居？
如果你只想检索数组边界内（包括对角线）所有单元格的邻居，以下代码就足够了。
public List<Element> getNeighbors( int x, int y ) {
List<Element> neighbors = new ArrayList<>();
for( int i = -1; i <= 1; ++i ) {
for( int j = -1; j <= 1; ++j ) {
if( i == 0 && j == 0 ) {
continue;
}
if( i + x >= 0 && i + x < array.length &&
j + y >= 0 && j + y < array[0].length ) {
// we found a valid neighbor!
neighbors.add( array[i][j] );
}
}
}
return neighbors;
}
public int[4][4] array2d;
//don't forget to fill it!
private void adjustNeighbors(int xCoord, int yCoord) {
for (int yi = y-1; yi <= yCoord+1; yi++) { //loop through the neighbors
for (int xi = x-1; xi <= xCoord+1; xi++) {
try {
if (!(xCoord != xi && yCoord != yi)) {
array2d[y][x]++; //do whatever you want to all the neighbors!
}
} catch (Exception e) {
// something is out of bounds
}
}
}
}
public void example(int changeSign, boolean shouldCheckRow,boolean shouldCheckColumn){
int num = 4;
if(changeSign < 0)
num = 0;
if(shouldCheckRow)
//adding a negative is the same as subtracting so if you add -1, you're really subtracting by one.
if(!((row + changeSign) < num))
//do stuff
else
if(!((col + changeSign) < num))
//do stuff
}
方法调用将会是：
public static void main(String args[]){
int shouldTestRight = 1;
int shouldTestLeft = -1;
int shouldTestUp = 1;
int shouldTestDown = -1;
// so if you want to test up or right, the first parameter should be positive
// if you want to test for down or left, the first parameter should be negative
// this is because the negative will flip the sign.
// if you should change the row, the second parameter should be true
// if you should change the column, the third parameter should be true.
example(shouldTestRight,true,false);
example(shouldTestLeft,true,false);
example(shouldTestUp,false,true);
example(shouldTestDown,false,true);
}
public class FindingNeighboursInMatrix {
public static void main(String[] args) {
int array[][] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 } };
for (int i = 0; i < array.length; i++) {
for (int j = 0; j < array[0].length; j++) {
System.out.println("neightbours of " + array[i][j]);
int neb[] = findneighbours(i, j, array);
for (int k = 0; k < neb.length; k++) {
if (neb[k] != -1) {
System.out.print(" " + neb[k] + ",");
}
}
System.out.println();
}
}
}
public static int[] findneighbours(int i, int j, int matrix[][]) {
int neb[] = new int[8];
// top row
neb[0] = getvalue(i - 1, j - 1, matrix);
neb[1] = getvalue(i - 1, j, matrix);
neb[2] = getvalue(i - 1, j + 1, matrix);
// left element
neb[3] = getvalue(i, j - 1, matrix);
// right element
neb[4] = getvalue(i, j + 1, matrix);
// bottom row
neb[5] = getvalue(i + 1, j - 1, matrix);
neb[6] = getvalue(i + 1, j, matrix);
neb[7] = getvalue(i + 1, j + 1, matrix);
return neb;
}
public static int getvalue(int i, int j, int matrix[][]) {
int rowSize = matrix.length;
int colSize = matrix[0].length;
if (i < 0 || j < 0 || i > rowSize - 1 || j > colSize - 1) {
return -1;
}
return matrix[i][j];
}}
//do stuff是什么。你能把它都改成一行吗？ - Christian Ternus!。例如，if(!((col-1)<0))可以改写为if((col-1)>=0)。 - ChiefTwoPencils