题目:

给定一个仅包含 0 和 1 、大小为 rows x cols 的二维二进制矩阵,找出只包含 1 的最大矩形,并返回其面积。

 

示例 1:

输入:matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]

输出:6

解释:最大矩形如上图所示。

示例 2:

输入:matrix = []

输出:0

示例 3:

输入:matrix = [["0"]]

输出:0

示例 4:

输入:matrix = [["1"]]

输出:1

示例 5:

输入:matrix = [["0","0"]]

输出:0

代码实现:

class Solution {
public int maximalRectangle(char[][] matrix) {
int m = matrix.length;
if (m == 0) {
return 0;
}
int n = matrix[0].length;
int[][] left = new int[m][n];

for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix[i][j] == '1') {
left[i][j] = (j == 0 ? 0 : left[i][j - 1]) + 1;
}
}
}

int ret = 0;
for (int j = 0; j < n; j++) { // 对于每一列,使用基于柱状图的方法
int[] up = new int[m];
int[] down = new int[m];

Deque<Integer> stack = new LinkedList<Integer>();
for (int i = 0; i < m; i++) {
while (!stack.isEmpty() && left[stack.peek()][j] >= left[i][j]) {
stack.pop();
}
up[i] = stack.isEmpty() ? -1 : stack.peek();
stack.push(i);
}
stack.clear();
for (int i = m - 1; i >= 0; i--) {
while (!stack.isEmpty() && left[stack.peek()][j] >= left[i][j]) {
stack.pop();
}
down[i] = stack.isEmpty() ? m : stack.peek();
stack.push(i);
}

for (int i = 0; i < m; i++) {
int height = down[i] - up[i] - 1;
int area = height * left[i][j];
ret = Math.max(ret, area);
}
}
return ret;
}
}