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   提示：

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 300
• matrix[i][j] 为 '0' 或 '1'

参考答案

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;
    }
};