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class Solution {//for each element in matrix, we compute h[j](current consecutive length of '1'), L[j] (nearest left wall whose height is smaller than current column), and R[j] (nearest right wall whose height is smaller than current column) for it, update these three terms row by row//here we compute the most far range(both in height, and width) of scan line j at every row i//to get the size of the rectangle of f[i][j]. To achieve this, we should compute://H[j](the maximum bottom up height of current column until meet the first '0')//L[j](the most far position where can the scan line j can go at the left)//R[j](the most far position where can the scan line j can go at the right)//then f[i][j]=H[j]*(R[j]-L[j]+1), ans=max(ans,f[i][j])public:	int maximalRectangle(vector
   
    
      > &matrix) {		// Start typing your C/C++ solution below		// DO NOT write int main() function		int ans = 0;		if(matrix.size() == 0) return ans;		int m = matrix[0].size();		vector
     
      L(m, -1);		vector
      
       R(m, m);		vector
       
        H(m, 0);		for (int i = 0; i < matrix.size(); ++i)		{			//scan from left to right to update H and L			int nearestLeft = -1;//virtual '0' at most far of right			for (int j = 0; j < matrix[i].size(); ++j)			{				L[j] = max(nearestLeft, L[j]);				if (matrix[i][j] == '1')					H[j]++;				else				{					H[j] = 0;					L[j] = -1;//note here					nearestLeft = j;				}			}			//scan from right to left to update R and calculate f[i][j]			int nearestRight = m;//virtual '0' at most far of left			for (int j = matrix[i].size()-1; j >= 0; --j)			{				R[j] = min(nearestRight, R[j]);				if (matrix[i][j] == '0')				{					nearestRight = j;					R[j] = m;//note here				}				//calculate f[i][j]				ans = max( ans, H[j]*(R[j]-L[j]-1) );			}					}		return ans;	}};
       
      
     
    
   

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