tag:
- Hard;
question:
Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.
Example:
Input:
[
["1","0","1","0","0"],
["1","0","1","1","1"],
["1","1","1","1","1"],
["1","0","0","1","0"]
]
Output: 6
思路：
此题是之前那道的Largest Rectangle in Histogram 的扩展，这道题的二维矩阵每一层向上都可以看做一个直方图，输入矩阵有多少行，就可以形成多少个直方图，对每个直方图都调用直方图中最大的矩形 中的方法，就可以得到最大的矩形面积。那么这道题唯一要做的就是将每一层构成直方图，由于题目限定了输入矩阵的字符只有 '0' 和 '1' 两种，所以处理起来也相对简单。方法是，对于每一个点，如果是‘0’，则赋0，如果是 ‘1’，就赋之前的height值加上1。具体参见代码如下：
class Solution {
public:
int maximalRectangle(vector<vector<char> > &matrix) {
int res = 0;
vector<int> height;
for (int i = 0; i < matrix.size(); ++i) {
height.resize(matrix[i].size());
for (int j = 0; j < matrix[i].size(); ++j) {
height[j] = matrix[i][j] == '0' ? 0 : (1 + height[j]);
}
res = max(res, largestRectangleArea(height));
}
return res;
}
int largestRectangleArea(vector<int> &height) {
int res = 0;
stack<int> s;
height.push_back(0);
for (int i = 0; i < height.size(); ++i) {
if (s.empty() || height[s.top()] <= height[i])
s.push(i);
else {
int tmp = s.top();
s.pop();
res = max(res, height[tmp] * (s.empty() ? i : (i - s.top() - 1)));
--i;
}
}
return res;
}
};
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