LeetCode Maximal Rectangle

最新推荐文章于 2019-04-26 14:33:48 发布

原创最新推荐文章于 2019-04-26 14:33:48 发布 · 513 阅读

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本文讨论了如何在给定的二维二进制矩阵中找到最大的矩形，并返回其面积。

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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.

Solution:

O(m*n),a combination of the problem:LeetCode Largest Rectangle in Histogram

class Solution {
public:
    int LargestRect(const vector<int>& height){
        vector<int> left;
        vector<int> right;
        stack<int> s;
        left.resize(height.size());
        right.resize(height.size());
        for(int i=0;i<height.size();++i){
            if(height[i]==0){
                s.push(i);
                continue;
            }
            while(!s.empty()&&height[s.top()]>=height[i])
                s.pop();
            left[i]=s.empty()?i:(i-s.top()-1);
            s.push(i);
        }
        s=stack<int>();
        for(int i=height.size()-1;i>=0;--i){
            if(height[i]==0){
                s.push(i);
                continue;
            }
            while(!s.empty()&&height[s.top()]>=height[i])
                s.pop();
            right[i]=s.empty()?height.size()-1-i:(s.top()-i-1);
            s.push(i);
        }
        int area=0;
        for(int i=0;i<height.size();++i){
            if(height[i]>0){
                area=max(area,height[i]*(left[i]+right[i]+1));
            }    
        }
        return area;
    }
    int maximalRectangle(vector<vector<char> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(matrix.empty())
            return 0;
        vector<int> height;
        int maxarea=0;
        height.resize(matrix[0].size());
        for(int i=0;i<matrix.size();++i){
            for(int j=0;j<matrix[i].size();++j){
                if(matrix[i][j]=='1'){
                    ++height[j];
                }else{
                    height[j]=0;
                }
            }
            maxarea=max(maxarea,LargestRect(height));
        }
        return maxarea;
    }
};