LeetCode_Maximal Rectangle
Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area.
分析: 对于每一列从右到左看成一个直方图,每个直方图计算最大面积的时间复杂度为O(n) 所以总的时间复杂度是O(n2)
class Solution {
public:
int maxArea(vector<int> &m)
{
int size = m.size();
stack<int> s;
int area, maxArea = ;
for(int i = ; i< size; ++i)
{
if(s.empty() || m[i] >= m[s.top()] )
{
s.push(i);
continue;
}
int tp = s.top();
s.pop();
area = m[tp] * (s.empty() ? i : i -s.top() -);
maxArea = maxArea > area ? maxArea : area;
--i;
} while(!s.empty()){
int tp = s.top();
s.pop();
area = m[tp] * (s.empty() ? size: size - s.top() -);
maxArea = maxArea > area ? maxArea : area;
}
return maxArea;
}
int maximalRectangle(vector<vector<char> > &matrix) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int row = matrix.size();
if(row < ) return ;
int column = matrix[].size();
if(column <) return ; int area , res = ; vector<int> m(row,);
for(int i = ; i< column; ++i)
{
for(int j = ; j< row; ++j)
if(matrix[j][i] == '')
m[j] = ;
else
m[j]++; area = maxArea(m);
res = res > area ? res : area;
}
return res;
}
};
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