[LeetCode] Largest Rectangle in Histogram、Maximal Rectangle

最新推荐文章于 2021-09-16 11:57:19 发布

原创最新推荐文章于 2021-09-16 11:57:19 发布 · 640 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

74 篇文章

订阅专栏

本文介绍了一种高效算法，用于解决给定直方图中寻找最大矩形面积的问题，并进一步扩展到二维矩阵中寻找全为1的最大矩形区域。通过使用单调栈实现，该算法能在O(n)的时间复杂度内解决问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Largest Rectangle in Histogram:

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.

For example,
Given height = [2,1,5,6,2,3],
return 10.

class Solution {
public:
    int largestRectangleArea(vector<int> &h) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        

        int n=h.size();
	    vector<int> lHigh(n,-1);
	    typedef pair<int,int> pos;
	    stack<pos> lQ;
	    lQ.push(pos(-1,-1));
	    for(int i=0;i<n;i++)
	    {
		    while(lQ.top().first>=h[i])
			    lQ.pop();
		    lHigh[i]=lQ.top().second+1;
		    lQ.push(pos(h[i],i));
	    }

	    vector<int> rHigh(n,-1);
	    stack<pos> rQ;
	    rQ.push(pos(-1,n));
	    for(int i=n-1;i>=0;i--)
	    {
		    while(rQ.top().first>=h[i])
			    rQ.pop();
		    rHigh[i]=rQ.top().second-1;
		    rQ.push(pos(h[i],i));
	    }

	    int res=0;
	    for(int i=0;i<n;i++)
		    res=max(res,(rHigh[i]-lHigh[i]+1)*h[i]);
	    return res;
    }
};

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

看来LeetCode把这两个题放在一起是有暗示的~

class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
    		int m=matrix.size();
			if ( m==0 )
					return 0;
			int n= matrix[0].size();

			vector<vector<char> > tmp(m,vector<char>(n,0));
            for(int i=0;i<n;i++)
                tmp[0][i]=matrix[0][i]-'0';
			for(int i=1;i<m;i++)
					for(int j=0;j<n;j++)
							if (matrix[i][j]=='1')
									tmp[i][j]=tmp[i-1][j]+matrix[i][j]-'0';
                            else
                                tmp[i][j]=0;
			int ret=0;
			for(int i=0;i<m;i++)
					ret=max(ret,solve(tmp[i]));
			return ret;
    }
	
	int solve(vector<char> &h) {
        int n=h.size();
	    vector<int> lHigh(n,-1);
	    typedef pair<int,int> pos;
	    stack<pos> lQ;
	    lQ.push(pos(-1,-1));
	    for(int i=0;i<n;i++)
	    {
		    while(lQ.top().first>=h[i])
			    lQ.pop();
		    lHigh[i]=lQ.top().second+1;
		    lQ.push(pos(h[i],i));
	    }

	    vector<int> rHigh(n,-1);
	    stack<pos> rQ;
	    rQ.push(pos(-1,n));
	    for(int i=n-1;i>=0;i--)
	    {
		    while(rQ.top().first>=h[i])
			    rQ.pop();
		    rHigh[i]=rQ.top().second-1;
		    rQ.push(pos(h[i],i));
	    }

	    int res=0;
	    for(int i=0;i<n;i++)
		    res=max(res,(rHigh[i]-lHigh[i]+1)*h[i]);
	    return res;
    }
    
};