刚才看见一段非常巧妙地算法来这里记录一下Maximum Rectangle

class Solution {

public:
int maximalRectangle(vector<vector<char> > &matrix) {
int row = matrix.size();
int col = matrix[0].size();

vector<vector<int> > trans(row, vector<int>(col, 0) );
for(int i = 0; i < row; i++) {
for(int j = 0; j < col; j++) {
if(matrix[i][i] == '0') {
trans[i][j] = 0;
}
else {
trans[i][j] = i == 0 ? 1 : trans[i - 1][j] + 1;
}
}
}
int maxArea = 0;
for(int i = 0; i < row; i++) {
maxArea = max(maxArea, largestRectangleArea(trans[i]));
}
return maxArea;
}

int largestRectangleArea(vector<int>& height) {
vector<int> tops;
int sum = 0;
int index = 0;

height.push_back(0);
while(index < height.size()) {
if(tops.empty() || height[index] > height[tops.back()]) {
tops.push_back(index);
index++;
}
else {
int cur = tops.back();
tops.pop_back();
int left = tops.back();
int right = index;

//sum = height[cur] * (top.empty() ? i : (right - left - 1) );
sum = max(sum, height[cur] * (tops.empty() ? right : (right - left - 1) ));
}
}
return sum;
}
};