class Solution { public: int maximalRectangle(vector<vector<char> > &matrix) { if (matrix.empty()) { return 0; } int n = matrix[0].size(); vector<int> H(n); vector<int> L(n); vector<int> 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。