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

这题自己写的算法复杂度很高，参考了leetcode官方讨论的算法，非常巧妙。