#yyds干货盘点# LeetCode程序员面试金典：最大正方形

题目：

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

代码实现：

class Solution {
    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;
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++) {
                if (matrix[i][j] == '1') {
                    // 遇到一个 1 作为正方形的左上角
                    maxSide = Math.max(maxSide, 1);
                    // 计算可能的最大正方形边长
                    int currentMaxSide = Math.min(rows - i, columns - j);
                    for (int k = 1; k < currentMaxSide; k++) {
                        // 判断新增的一行一列是否均为 1
                        boolean flag = true;
                        if (matrix[i + k][j + k] == '0') {
                            break;
                        }
                        for (int m = 0; m < k; m++) {
                            if (matrix[i + k][j + m] == '0' || matrix[i + m][j + k] == '0') {
                                flag = false;
                                break;
                            }
                        }
                        if (flag) {
                            maxSide = Math.max(maxSide, k + 1);
                        } else {
                            break;
                        }
                    }
                }
            }
        }
        int maxSquare = maxSide * maxSide;
        return maxSquare;
    }
}