标签：har || each etc nta container led not ret
To appy DP, we define the state as the maximal size (square = size * size) of the square that can be formed till point (i, j), denoted as dp[i][j].
For the topmost row (i = 0) and the leftmost column (j = 0), we have dp[i][j] = matrix[i][j] - ‘0‘, meaning that it can at most form a square of size 1 when the matrix has a ‘1‘in that cell.
When i > 0 and j > 0, if matrix[i][j] = ‘0‘, then dp[i][j] = 0 since no square will be able to contain the ‘0‘ at that cell. If matrix[i][j] = ‘1‘, we will have dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1, which means that the square will be limited by its left, upper and upper-left neighbors.
class Solution { public: int maximalSquare(vector<vector<char>>& matrix) { if (matrix.empty()) { return 0; } int m = matrix.size(), n = matrix[0].size(), sz = 0; vector<vector<int>> dp(m, vector<int>(n, 0)); for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (!i || !j || matrix[i][j] == ‘0‘) { dp[i][j] = matrix[i][j] - ‘0‘; } else { dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1; } sz = max(dp[i][j], sz); } } return sz * sz; } };
In the above code, it uses O(mn) space. Actually each time when we update dp[i][j], we only need dp[i-1][j-1], dp[i-1][j] (the previous row) and dp[i][j-1] (the current row). So we may just keep two rows.
class Solution { public: int maximalSquare(vector<vector<char>>& matrix) { if (matrix.empty()) { return 0; } int m = matrix.size(), n = matrix[0].size(), sz = 0; vector<int> pre(n, 0), cur(n, 0); for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (!i || !j || matrix[i][j] == ‘0‘) { cur[j] = matrix[i][j] - ‘0‘; } else { cur[j] = min(pre[j - 1], min(pre[j], cur[j - 1])) + 1; } sz = max(cur[j], sz); } fill(pre.begin(), pre.end(), 0); swap(pre, cur); } return sz * sz; } };
Furthermore, we may only use just one vector (thanks to @stellari for sharing the idea).
class Solution { public: int maximalSquare(vector<vector<char>>& matrix) { if (matrix.empty()) { return 0; } int m = matrix.size(), n = matrix[0].size(), sz = 0, pre; vector<int> cur(n, 0); for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { int temp = cur[j]; if (!i || !j || matrix[i][j] == ‘0‘) { cur[j] = matrix[i][j] - ‘0‘; } else { cur[j] = min(pre, min(cur[j], cur[j - 1])) + 1; } sz = max(cur[j], sz); pre = temp; } } return sz * sz; } };
标签：har || each etc nta container led not ret
原文地址：https://www.cnblogs.com/shona/p/11311161.html
