【LeetCode】85. Maximal Rectangle

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.

如果用DP来做，判断(begin1,end1)~(begin2,end2)范围是否全1，会超时。

对于矩阵matrix，逐行构建height数组，调用Largest Rectangle in Histogram即可。

对matrix第i行构建height数组方法：对于height[j]，即从matrix[i][j]开始，最多到达matrix[0][j]的连续1个数。

class Solution {
public:

    int maximalRectangle(vector<vector<char> > &matrix) {
        int ret = ;;
        if(matrix.empty() || matrix[].empty())
            return ret;
        int m = matrix.size();
        int n = matrix[].size();
        for(int i = ; i < matrix.size(); i ++)
        {
            vector<int> height(n, );
            for(int j = ; j < n; j ++)
            {
                int r = i;
                while(r >=  && matrix[r][j] == '')
                {
                    height[j] ++;
                    r --;
                }
            }
            ret = max(ret, largestRectangleArea(height));
        }
        return ret;
    }

    int largestRectangleArea(vector<int> &height) {
        if(height.empty())
            return ;

        int result = ;
        stack<int> s;   //elements in stack s are kept in ascending order
        int ind = ;
        while(ind < height.size())
        {
            if(s.empty() || height[ind]>=s.top())
            {
                s.push(height[ind]);
            }
            else
            {
                int count = ;  //pop count
                while(!s.empty() && height[ind]<s.top())
                {
                    int top = s.top();
                    s.pop();
                    count ++;
                    result = max(result, count*top);
                }
                for(int i = ; i <= count; i ++)
                    s.push(height[ind]);    //push count+1 times
            }
            ind ++;
        }
        //all elements are in stack s, and in ascending order
        int count = ;
        while(!s.empty())
        {
            count ++;
            int top = s.top();
            s.pop();
            result = max(result, count*top);
        }

        return result;
    }
};