我正在尝试以螺旋顺序填充数组。到目前为止，我可以按螺旋顺序打印数组，但有没有办法修改数组，以便我可以按螺旋顺序填充它然后只打印数组？我希望它像倒计时一样按顺序递减。请帮忙！
public class Spiral {
public static void main(int m, int n) {
// create m by n array of integers 1 through m*n
int[][] values = new int[m][n];
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
values[i][j] = 1 + (m*n)*i + j;
// spiral
for (int i = (m*n)-1, j = 0; i > 0; i--, j++) {
for (int k = j; k < i; k++) System.out.println(values[j][k]);
for (int k = j; k < i; k++) System.out.println(values[k][i]);
for (int k = i; k > j; k--) System.out.println(values[i][k]);
for (int k = i; k > j; k--) System.out.println(values[k][j]);
}
}
}
答案 0 :(得分：1)
int maxValue = target.length * target[0].length;
private int[][] generateMatrix(int[][] target, int level, int currentVal) {
// always start from lower left corner in each layer
int w = level;
int h = target.length - level - 1;
// fill the bottom line
int i = 0;
for (i = w; i < target[0].length - level && currentVal <= maxValue; i++) {
target[h][i] = currentVal++;
}
w = target[0].length - level - 1;
int j = 0;
// fill the right line
for (j = h - 1; j >= level && currentVal <= maxValue; j--) {
target[j][w] = currentVal++;
}
h = level;
// fill the above line
for (i = w - 1; i >= level && currentVal <= maxValue; i--) {
target[h][i] = currentVal++;
}
w = level;
// fill the left line
for (j = h + 1; j < target.length - level - 1 && currentVal <= maxValue; j++) {
target[j][w] = currentVal++;
}
if (currentVal > maxValue)
return target;
return generateMatrix(target, ++level, currentVal);
}
答案 1 :(得分：1)
下面的函数是一个大小为N×N的方矩阵，其中包含从1到N * N的螺旋形整数，从左上角开始沿顺时针方向旋转。
int[][] spiralNumbers(int n) {
int[][] matrix = new int[n][n];
for (int step = 0, a = 0, size; step < n/2; step++) {
size = (n - step * 2 - 1);
for (int i = 0, chunk, chunkIndex, chunkOffset; i < 4 * size; i++) {
chunk = i / size;
chunkIndex = i % size;
chunkOffset = n - step - 1;
switch (chunk) {
case 0:
matrix[step][chunkIndex + step] = a+1;
break;
case 1:
matrix[chunkIndex + step][chunkOffset] = a+1;
break;
case 2:
matrix[chunkOffset][chunkOffset - chunkIndex] = a+1;
break;
case 3:
matrix[chunkOffset - chunkIndex][step] = a+1;
break;
default:
throw new IndexOutOfBoundsException();
}
a++;
}
if (n % 2 == 1) {
matrix[n/2][n/2] = n * n;
}
}
return matrix;
}
答案 2 :(得分：0)
如果您已经找到了执行读取的代码（用于打印），那么您肯定可以使用相同的逻辑来修改它以进行写入吗？
如果您希望矩阵中的每个单元格都包含其“序列号”，向后计数，这样的事情应该有效，假设您的访问逻辑是正确的：
for (int i = (m*n)-1, j = 0, index = m * n; i > 0; i--, j++) {
for (int k = j; k < i; k++) values[j][j] = index--;
for (int k = j; k < i; k++) values[k][i] = index--;
for (int k = i; k > j; k--) values[i][k] = index--;
for (int k = i; k > j; k--) values[k][j] = index--;
}
答案 3 :(得分：0)
效率不高，但应该有效：g是数组。我也使用异常来控制逻辑。
public static void spiralFill()
{
x = (g.length-1)/2;
y = (g[0].length-1)/2;
try
{
while(true)
{
east();
south();
step++;
west();
north();
step++;
}
}
catch(ArrayIndexOutOfBoundsException e)
{
}
}
public static void east()
{
for(int i = 0; i < step; i++)
{
g[x][y] = count;
count++;
x++;
}
}
public static void south()
{
for(int i = 0; i < step; i++)
{
g[x][y] = count;
count++;
y--;
}
}
public static void west()
{
for(int i = 0; i < step; i++)
{
g[x][y] = count;
count++;
x--;
}
}
public static void north()
{
for(int i = 0; i < step; i++)
{
g[x][y] = count;
count++;
y++;
}
}
答案 4 :(得分：0)
以下是一些变体：
public class SpiralMatrix {
private static int[][] createSpiralMatrix(int size) {
int[][] matrix = new int[size][size];
int row = 0, col = -1;
int value = 1;
boolean horizontal = true;
boolean increasing = true;
boolean finish = false;
while(!finish) {
finish = true;
if (horizontal && increasing) {
while(tryAndSet(matrix, row, col + 1, value)) {
finish = false;
col++;
value++;
}
} else if (horizontal && !increasing) {
while(tryAndSet(matrix, row, col - 1, value)) {
finish = false;
col--;
value++;
}
} else if (!horizontal && increasing) {
while(tryAndSet(matrix, row + 1, col, value)) {
finish = false;
row++;
value++;
}
} else {
while(tryAndSet(matrix, row - 1, col, value)) {
finish = false;
row--;
value++;
}
}
if (!horizontal) {
increasing = !increasing;
}
horizontal = !horizontal;
}
return matrix;
}
private static boolean tryAndSet(int[][] matrix, int row, int col, int value) {
if (row < 0 || col < 0 || row >= matrix.length || col >= matrix[row].length || matrix[row][col] != 0) {
return false;
}
matrix[row][col] = value;
return true;
}
private static void printMatrix(int[][] matrix) {
for (int i = 0; i < matrix.length; i++) {
for(int j = 0; j < matrix[i].length; j++) {
System.out.print("\t" + matrix[i][j]);
}
System.out.println();
}
}
public static void main(String[] args) {
try {
int[][] spiralMatrix = createSpiralMatrix(40);
printMatrix(spiralMatrix);
} catch (Throwable th) {
th.printStackTrace();
}
}
答案 5 :(得分：0)
public ArrayList<Integer> spiralOrder(final List<ArrayList<Integer>> a) {
ArrayList<Integer> result = new ArrayList<Integer>();
int m = a.get(0).size();
int n = a.size();
if(m>1 && n>1){
int loopCounter = (n > m) ? m*2 : n *2 -1 ;
int opr=1;
int i=0,j=0;
int opA=m,opB=n,opC=0,opD=1;
for(int k=0;k < loopCounter ;k++){
if(opr == 1){
int counter =0;
while(counter < opA){
System.out.print(a.get(i).get(j)+ ";");
result.add(a.get(i).get(j));
counter++;
if(j != opA-1){
j++;
}
else{
break;
}
}
opr =2;
continue;
}
if(opr == 2){
i++;
int counter =1;
while(counter < opB){
System.out.print(a.get(i).get(j)+ ";");
result.add(a.get(i).get(j));
counter++;
if( i != opB-1){
i++;
}
else{
break;
}
}
opr =3;
continue;
}
if(opr == 3){
j--;
int counter =j;
while(counter >= opC){
System.out.print(a.get(i).get(j)+ ";");
result.add(a.get(i).get(j));
counter --;
if(j != opC){
j--;
}
else{
break;
}
}
opr =4;
continue;
}
if(opr == 4){
i--;
int counter = i;
while(counter >= opD){
System.out.print(a.get(i).get(j)+ ";");
result.add(a.get(i).get(j));
counter --;
if(i != opD){
i--;
}else{
break;
}
}
opr =1;
j++;
opA = opA -1;
opB = opB -1;
opC= opC +1;
opD = opD+1;
continue;
}
}
}
else if(n ==1){
for(int k=0;k < a.get(0).size();k++){
result.add(a.get(0).get(k));
}
}
else if(m==1 && n==1){
result.add(a.get(0).get(0));
}
// Populate result;
return result;
}