力扣85.最大矩形

题目：biubiu
题意：给一个二维矩阵，存的是1和0，求1可以构成的最大矩形。
思维转换，力扣84柱状图的最大矩形的变换。

class Solution {
    
    
public:
	int largestRectangleArea(vector<int>& heights) {
    
    
		int n = heights.size();
		vector<int>l(n), r(n,n);
		stack<int>p;
		for (int i = 0; i < n; i++) {
    
    
			while (!p.empty() && heights[p.top()] >= heights[i]) {
    
    
				r[p.top()] = i;
				   p.pop();
				}
					
				l[i] = p.empty() ? -1 : p.top();
				p.push(i);	
		}
		/*while (p.size())
			p.pop();
		for (int i = n - 1; i >= 0; i--) {
			while (!p.empty() && heights[p.top()] >= heights[i])
				p.pop();
			r[i] = p.empty() ? n : p.top();
			p.push(i);
		}*/
		int ans = 0;
		for (int i = 0; i < n; i++) {
    
    
			ans = max(ans, (r[i] - l[i] - 1) * heights[i]);
		}
		return ans;
	}
	int maximalRectangle(vector<vector<char>>& matrix) {
    
    
		int ans = 0;
        if(matrix.size()==0)
           return 0;
		vector<int>heights(matrix[0].size(),0);
		for (const auto& p:matrix) {
    
    
			for (int i = 0;i<p.size();i++) {
    
    
				if (p[i] == '1')
					heights[i]++;
                else
                    heights[i]=0;
			}
            /*for(const auto&q:heights){
                cout<<q<<" ";
            }
            cout<<endl;*/
            //cout<<largestRectangleArea(heights)<<endl;
			ans = max(ans, largestRectangleArea(heights));
		}
		return ans;
	}
};

方法二：

class Solution {
    
    
public:
	int maximalRectangle(vector<vector<char>>& matrix) {
    
    
		if (matrix.size() == 0)
			return 0;
		vector<vector<int>>widths(matrix.size(), vector<int>(matrix[0].size(),0));
		for (int i = 0;i<matrix.size();i++) {
    
    
			for (int j = 0; j < matrix[i].size(); j++) {
    
    
				if (matrix[i][j] == '1')
					widths[i][j] = (j == 0 ? 1 : widths[i][j - 1] + 1);
			}
		}
		int ans = 0;
		for (int i = 0; i < matrix.size(); i++) {
    
    
			for (int j = 0; j < matrix[i].size(); j++) {
    
    
				if (matrix[i][j] == '0')
					continue;
				int h = j + 2;
				for (int k = i; k >= 0; k--) {
    
    
					h = min(h, widths[k][j]);
					ans = max(ans, (i-k + 1)*h);
				}
			}
		}
		return ans;
	}
};

官方题解思路，先遍历找到每一行的可以形成矩形的长，然后在遍历得到高。