Medium | LeetCode 221. 最大正方形 | 动态规划

221. 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

提示：

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

动态规划

状态转移方程：

\[d p(i, j)=\min (d p(i-1, j), d p(i-1, j-1), d p(i, j-1))+1 \]

public int maximalSquare(char[][] matrix) {
    int maxSide = 0;
    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
        return maxSide;
    }
    int rows = matrix.length, columns = matrix[0].length;
    int[][] dp = new int[rows][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] = Math.min(
                        Math.min(dp[i - 1][j], dp[i][j - 1]), 
                        dp[i - 1][j - 1]) + 1;
                }
                maxSide = Math.max(maxSide, dp[i][j]);
            }
        }
    }
    int maxSquare = maxSide * maxSide;
    return maxSquare;
}