leetcode之Largest Rectangle in Histogram、Maximal Rectanglex；待字闺中之最大矩形分析

题目一：Largest Rectangle in Histogram 

参考博客

int largestRectangleArea(vector<int> &height){
	height.push_back(0);//加入一个最小高度，用于清空最后栈中的所有数据
	int length = height.size(),i = 0,maxArea = 0;
	stack<int> stackIndex;
	while(i < length)
	{
		if(stackIndex.empty() || height[i] >= height[stackIndex.top()])stackIndex.push(i++);//高度比前面的大，则进栈
		else
		{
			int index = stackIndex.top();//弹出栈顶元素，并计算高度
			stackIndex.pop();
			int area = height[index] * (stackIndex.empty() ? i : (i - stackIndex.top() - 1));
			maxArea = max(maxArea,area);
		}
	}
	return maxArea;
}

题目二：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.

参考博客

int largestRectangleArea(vector<int> &height){
	height.push_back(0);
	int length = height.size(),i = 0,maxArea = 0;
	stack<int> stackIndex;
	while(i < length)
	{
		if(stackIndex.empty() || height[i] >= height[stackIndex.top()])stackIndex.push(i++);
		else
		{
			int index = stackIndex.top();
			stackIndex.pop();
			int area = height[index] * (stackIndex.empty() ? i : (i - stackIndex.top() - 1));
			maxArea = max(maxArea,area);
		}
	}
	return maxArea;
}
int maximalRectangle(vector<vector<char> > &matrix) {
	int row = matrix.size();
	int col = matrix[0].size();
	int i,j;
	vector<vector<int> >dp(row);//dp[i][j]表示从第i个元素开始，包括第i个元素，向上数，直到遇到0时，1的个数,参考该博客
	for(i = 0;i < row;i++)
	{
		vector<int> tmp(col);
		dp[i] = tmp;
	}
	for(j = 0;j < col;j++)dp[0][j] = matrix[0][j] - '0';
	for(i = 1;i < row;i++)
	{
		for(j=0;j < col;j++)
		{
			if(matrix[i][j] == '0')dp[i][j] = 0;
			else dp[i][j] = dp[i-1][j] + 1;
		}
	}
	int maxArea = 0;
	for(i = 0;i < row;i++)
	{
		maxArea = max(maxArea,largestRectangleArea(dp[i]));
	}
	return maxArea;
}

由于参考的博客写的非常好，自己就不想多写了，记录下来供自己以后回顾，希望