http://www.hawstein.com/posts/20.11.html
Imagine you have a square matrix, where each cell is filled with either black or white. Design an algorithm to find the maximum subsquare such that all four borders are filled with black pixels.
译文：
有一个正方形矩阵，里面的每一个小格子要么被涂上黑色要么被涂上白色。 设计算法，找到四条边都是黑色格子的最大子正方形。
解答
暴力法，从左到右，从上到下遍历格子，将它作为子正方形左上角的点。 固定了子正方形左上角的点，我们只需要知道边长，就能把子正方形确定下来。 我们按边长从大到小开始，去检查每一个子正方形的四条边是否都为黑色格子。 如果是，则记下当前最大的边长值。将子正方形左上角的点移动到下一行(即向下移动一格)， 进入下一轮循环。
public class SquareCell {
public int zerosRight = 0;
public int zerosBelow = 0;
public SquareCell(int right, int below) {
zerosRight = right;
zerosBelow = below;
}
}
public class Subsquare {
public int row, column, size;
public Subsquare(int r, int c, int sz) {
row = r;
column = c;
size = sz;
}
}
public static Subsquare findSquareWithSize(SquareCell[][] processed, int square_size) {
// On an edge of length N, there are (N - sz + 1) squares of length sz.
int count = processed.length - square_size + 1;
// Iterate through all squares with side length square_size.
for (int row = 0; row < count; row++) {
for (int col = 0; col < count; col++) {
if (isSquare(processed, row, col, square_size)) {
return new Subsquare(row, col, square_size);
}
}
}
return null;
}
public static Subsquare findSquare(int[][] matrix){
assert(matrix.length > 0);
for (int row = 0; row < matrix.length; row++){
assert(matrix[row].length == matrix.length);
}
SquareCell[][] processed = processSquare(matrix);
int N = matrix.length;
for (int i = N; i >= 1; i--) {
Subsquare square = findSquareWithSize(processed, i);
if (square != null) {
return square;
}
}
return null;
}
private static boolean isSquare(SquareCell[][] matrix, int row, int col, int size) {
SquareCell topLeft = matrix[row][col];
SquareCell topRight = matrix[row][col + size - 1];
SquareCell bottomRight = matrix[row + size - 1][col];
if (topLeft.zerosRight < size) { // Check top edge
return false;
}
if (topLeft.zerosBelow < size) { // Check left edge
return false;
}
if (topRight.zerosBelow < size) { // Check right edge
return false;
}
if (bottomRight.zerosRight < size) { // Check bottom edge
return false;
}
return true;
}
public static SquareCell[][] processSquare(int[][] matrix) {
SquareCell[][] processed = new SquareCell[matrix.length][matrix.length];
for (int r = matrix.length - 1; r >= 0; r--) {
for (int c = matrix.length - 1; c >= 0; c--) {
int rightZeros = 0;
int belowZeros = 0;
if (matrix[r][c] == 0) { // only need to process if it's a black cell
rightZeros++;
belowZeros++;
if (c + 1 < matrix.length) { // next column over is on same row
SquareCell previous = processed[r][c + 1];
rightZeros += previous.zerosRight;
}
if (r + 1 < matrix.length) {
SquareCell previous = processed[r + 1][c];
belowZeros += previous.zerosBelow;
}
}
processed[r][c] = new SquareCell(rightZeros, belowZeros);
}
}
return processed;
}
public static Subsquare findSquareWithSize(int[][] matrix, int squareSize) {
// On an edge of length N, there are (N - sz + 1) squares of length sz.
int count = matrix.length - squareSize + 1;
// Iterate through all squares with side length square_size.
for (int row = 0; row < count; row++) {
for (int col = 0; col < count; col++) {
if (isSquare(matrix, row, col, squareSize)) {
return new Subsquare(row, col, squareSize);
}
}
}
return null;
}
public static Subsquare findSquare(int[][] matrix){
assert(matrix.length > 0);
for (int row = 0; row < matrix.length; row++){
assert(matrix[row].length == matrix.length);
}
int N = matrix.length;
for (int i = N; i >= 1; i--) {
Subsquare square = findSquareWithSize(matrix, i);
if (square != null) {
return square;
}
}
return null;
}
private static boolean isSquare(int[][] matrix, int row, int col, int size) {
// Check top and bottom border.
for (int j = 0; j < size; j++){
if (matrix[row][col+j] == 1) {
return false;
}
if (matrix[row+size-1][col+j] == 1) {
return false;
}
}
// Check left and right border.
for (int i = 1; i < size - 1; i++) {
if (matrix[row+i][col] == 1){
return false;
}
if (matrix[row+i][col+size-1] == 1) {
return false;
}
}
return true;
}
