LeetCode Maximal Rectangle

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.


O(N^3)

class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
    	//check constraint
		int x = matrix.size();
		if(0 == x)	return 0;
		int y = matrix[0].size();
		if(0 == y)	return 0;

		vector<vector<int> > result(x, vector<int>(y));
		for(int i = 0; i < x; ++i)
			for(int j = 0; j < y; ++j)
				result[i][j] = '0' == matrix[i][j]? 0: 1;

		int ret = 0;
		//transform the matrix line by line. O(N^2)
		//matrix[i][j] is number of consecutive 1s counting upward
		for(int i = 1; i < x; ++i)
			for(int j = 0; j < y; ++j)
				result[i][j] += ((0 == result[i][j])? 0: result[i-1][j]);
		//for each line
		for(int i = 0; i < x; ++i)
		{
			int maxHeight = result[i][0];
			//for each column, keep the min height, calculate the sum
			for(int start = 0; start < y; ++start)
			{
				int maxHeight = result[i][start];
				for(int end = start; end < y; ++end)
				{
					//maxHeight = min(maxHeight, matrix[i][end]);
					maxHeight = min(maxHeight, result[i][end]);
					ret = max(ret, maxHeight*(end - start + 1));
				}
			}
		}
		return ret;
    }
};

O(N^2)
class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
    	int x = matrix.size();
		if(0 == x)	return 0;
		int y = matrix[0].size();
		if(0 == y)	return 0;

		vector<vector<int> > result(x, vector<int>(y));
		for(int i = 0; i < x; ++i)
			for(int j = 0; j < y; ++j)
				result[i][j] = '0' == matrix[i][j]? 0: 1;
		for(int i = 1; i < x; ++i)
			for(int j = 0; j < y; ++j)
				result[i][j] += 0 == result[i][j]? 0: result[i-1][j];
		int ret = 0;
		for(int i = 0; i < x; ++i)
			ret = max(ret, maxArea(result[i]));
		return ret;
    }

	int maxArea(vector<int>& line)
	{
		stack<int> st;
		vector<int> leftLimit(line.size());
		for(int i = 0; i < line.size(); ++i)
		{
			while(!st.empty())
			{
				if(line[st.top()] >= line[i])
					st.pop();
				else
					break;
			}
			leftLimit[i] = st.empty()? -1: st.top();
			st.push(i);
		}
		while(!st.empty())	st.pop();
		vector<int> rightLimit(line.size());
		for(int i = line.size() - 1; i >= 0; --i)
		{
			while(!st.empty())
			{
				if(line[st.top()] >= line[i])
					st.pop();
				else
					break;
			}
			rightLimit[i] = st.empty()? line.size(): st.top();
			st.push(i);
		}
		int ret = 0;
		for(int i = 0; i < line.size(); ++i)
			ret = max(ret, line[i]*(rightLimit[i] - leftLimit[i] - 1));
		return ret;
	}
};


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