Given a 2D binary matrix filled with 0’s and 1’s, find the largest square containing all 1’s and return its area.
For example, given the following matrix:
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| 1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
|
Return 4.
To finish this, please implement this first: Maximal Rectangle
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| public class MaximalSquare {
public int maximalSquare(char[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return 0;
}
int h = matrix.length;
int w = matrix[0].length;
int[][] map = new int[h][w + 1];
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
if (i == 0) {
map[i][j] = matrix[i][j] - '0';
} else {
if (matrix[i][j] == '0') {
map[i][j] = 0;
} else {
map[i][j] = 1 + map[i - 1][j];
}
}
}
}
int max = 0;
for (int i = 0; i < h; i++) {
max = Math.max(max, helper(map[i]));
}
return max;
}
private int helper(int[] height) {
if (height == null || height.length == 0) {
return 0;
}
Stack<Integer> stack = new Stack<Integer>();
int crt, max;
crt = max = 0;
while (crt < height.length) {
if (stack.isEmpty() || height[crt] >= height[stack.peek()]) {
stack.push(crt++);
} else {
int top = stack.pop();
int length = Math.min(height[top], (stack.isEmpty() ? crt : crt - stack.peek() - 1));
max = Math.max(max, length * length);
}
}
return max;
}
}
|