最大正方形（二维dp）

最大正方形
在一个由 '0' 和 '1' 组成的二维矩阵内，找到只包含 '1' 的最大正方形，并返回其面积。

示例 1：

输入：matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
输出：4
示例 2：

输入：matrix = [["0","1"],["1","0"]]
输出：1
示例 3：

输入：matrix = [["0"]]
输出：0

题解

可以使用动态规划降低时间复杂度。我们用 \(dp[i][j]\)表示以下标 \(i\) 和 \(j\) 为全部包含“1”的正方形的右下角。则当\(matrix[i][j]==1\)时有状态转移方程：\(dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1])+1\)
当\(matrix[i][j]==0\)时，\(dp[i][j]=0\)
最后，还需考虑边界条件，即当\(i=0\) 或 \(j=0\)时，若 \(matrix[i][j]=1\) ,则 \(dp[i][j]=1\),否则\(dp[i][j]=0\)

class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        if (matrix.size() == 0 || matrix[0].size() == 0) {
            return 0;
        }
        int maxSide = 0;
        int rows = matrix.size(), columns = matrix[0].size();
        vector<vector<int>> dp(rows, vector<int>(columns));
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++) {
                if (matrix[i][j] == '1') {
                    if (i == 0 || j == 0) {
                        dp[i][j] = 1;
                    } else {
                        dp[i][j] = min(min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1;
                    }
                    maxSide = max(maxSide, dp[i][j]);
                }
            }
        }
        int maxSquare = maxSide * maxSide;
        return maxSquare;
    }
};