【LeetCode】Maximal Rectangle_【leetcode】maximal rectangle (dp)-优快云博客

本文链接：https://blog.youkuaiyun.com/snowwolf_yang/article/details/37384767

该博客主要探讨了LeetCode上的Maximal Rectangle问题，即在一个二维的二进制矩阵中找到最大的全1矩形并返回其面积。博主提供了详细的题目描述、分析，以及C++实现的解决方案，包括一个名为Solution的类，通过维护高度(H)、左边界(L)和右边界(R)数组来计算最大矩形的面积。

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参考链接

http://blog.youkuaiyun.com/littlestream9527/article/details/19641013

http://blog.youkuaiyun.com/nandawys/article/details/9245119

http://www.cnblogs.com/yingzhongwen/archive/2013/04/26/3044560.html

http://wenku.baidu.com/link?url=Exq42d-rpqmRGziQGTsUvDmAFshiuXAt-QIbPwMebDRkDm4s9AfmO9nOZKfmX4G-sDJ-ljNp1jde7ibh_8z-0bBubDo34MxSp1_cQcZRsAm

题目描述

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.

题目分析

这个题目有两种经典的解法。

一：转换成柱形图求最大面积。n^2

二：浅谈用极大化思想解决最大子矩形问题

总结

代码示例

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> S;
        line.push_back(0);
        int sum = 0;
        for (int i = 0; i < line.size(); i++) {
            if (S.empty() || line[i] > line[S.top()]) S.push(i);
            else {
                int tmp = S.top();
                S.pop();
                sum = max(sum, line[tmp]*(S.empty()? i : i-S.top()-1));
                i--;
            }
        }
        return sum;
    }
};

class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix){
		if(matrix.size()==0)
			return 0;
		int n = matrix[0].size();
		vector<int> H(n);
		vector<int> L(n);
		vector<int> R(n,n);
		int res = 0;
		for(int i=0;i<matrix.size();++i){
			int left = 0, right = n;
			for(int j=0;j<n;++j){
				if(matrix[i][j]=='1'){
					++H[j];
					L[j] = max(left,L[j]);
				}
				else{
					H[j] = 0; left = j+1; L[j] = 0; R[j] = n;
				}
			}
			for(int j=n-1;j>=0;--j){
				if(matrix[i][j]=='1'){
					R[j] = min(R[j],right);
					res = max(res, (R[j]-L[j])*H[j]);
				}
				else{
					right = j;
				}
			}
		}
		return res;

	}
};