LeetCode(85) Maximal Rectangle

Posted 2022-06-22 逆風的薔薇

题目

Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 6.

分析

求给定只包含'0'和‘1’的矩阵中，由连续的‘1’组成的最大矩阵面积。这道题基于直方图的最大长方形的延伸，听过左神在牛客上的算法精讲课的小伙伴应该都有很清晰的思路。 这道题的关键就是 先将矩阵转化为数组，算法原型即求直方图的最大面积。

代码

/*
	LeetCode 85_Maximal Rectangle.cpp
*/

#include <iostream>
#include <cstdlib>
#include <vector>
#include <stack>
#include <algorithm>

using namespace std;



class Solution 
public:
	int maximalRectangle(vector<vector<char>>& matrix) 
		if (matrix.empty())
			return 0;

		int m = matrix.size(), n = matrix[0].size();
		vector<int> heights(n, 0);
		int maxArea = 0;
		for (int i = 0; i < m; ++i)
		
			for (int j = 0; j < n; ++j)
			
				if (matrix[i][j] == '1')
				
					++heights[j];
				else
					heights[j] = 0;
				//else			
			//for
			maxArea = max(maxArea, largestRectangleArea(heights));
		//for
		return maxArea;
	

	/*求直方图中最大矩形面积，利用栈*/
	int largestRectangleArea(vector<int> &heights)
 	
		if (heights.empty())
			return 0;

		int n = heights.size();
		int maxArea = 0;
		stack<int> st;
		for (int i = 0; i < n; ++i)
		
			/*当前高度高于栈顶索引*/
			while (!st.empty() && heights[i] <= heights[st.top()])
			
				//计算栈顶索引左右扩展对应的最大长方形面积
				int idx = st.top();
				st.pop();
				int left = st.empty() ? -1 : st.top();
				int curArea = (i - left - 1)*heights[idx];

				maxArea = max(maxArea, curArea);
			//if
			st.push(i);
		//for

		/*检验栈中剩余元素*/
		while (!st.empty())
		
			int idx = st.top();
			st.pop();

			int left = st.empty() ? -1 : st.top();
			int curArea = (n - left - 1)*heights[idx];

			maxArea = max(maxArea, curArea);
		//while

		return maxArea;
	
;

int main()

	vector<vector<char>> v = '1','0','1','1','1',
							'0','1','0','1','0',
							 '1','1','0','1','1' ,
							 '0','1','1','1','1' 
	;

	cout << Solution().maximalRectangle(v) << endl;

	system("pause");
	return 0;