Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.
Input:
[
["1","0","1","0","0"],
["1","0","1","1","1"],
["1","1","1","1","1"],
["1","0","0","1","0"]
]
Output: 6
["1","0","1","0","0"] --> 0
["2","0","2","1","1"] --> 3*1
["3","1","3","2","2"] --> 3*2
["4","1","3","3","2"] --> 5*1
class Solution {
public int maximalRectangle(char[][] matrix) {
int res = 0;
if (matrix == null || matrix.length == 0 ||
matrix[0] == null || matrix[0].length == 0) {
return res;
}
int m = matrix.length;
int n = matrix[0].length;
int[] height = new int[n + 1];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix[i][j] == '1') {
height[j] += 1;
} else {
height[j] = 0;
}
}
res = Math.max(res, hist(height));
}
return res;
}
public int hist(int[] height) {
int res = 0;
Stack<Integer> stack = new Stack<>();
for (int i = 0; i < height.length; i++) {
while (!stack.isEmpty() && height[i] <= height[stack.peek()]) {
int h = height[stack.pop()];
int w = stack.isEmpty() ? i : i - stack.peek() - 1;
res = Math.max(res, h * w);
}
stack.push(i);
}
return res;
}
}