链接

求解

1、暴力法

2、动态规划 DP

/**
 * 方法二：动态规划
 * DP
 */
class Solution2 {
    public int maximalSquare(char[][] matrix) {
        int rows = matrix.length, cols = rows > 0 ? matrix[0].length : 0;
        // 维数与原矩阵一致 dp[i][j]表示的是由当前元素作为边长时，组成的最大正方形的边长
        int[][] dp = new int[rows + 1][cols + 1];
        int maxsqlen = 0;
        for (int i = 1; i <= rows; i++) {
            for (int j = 1; j <= cols; j++) {
                if (matrix[i-1][j-1] == '1'){
                    dp[i][j] = Math.min(Math.min(dp[i][j-1], dp[i-1][j]), dp[i-1][j-1]) + 1;
                    maxsqlen = Math.max(dp[i][j],maxsqlen);
                }
            }  
        }
        return maxsqlen * maxsqlen;
    }
    public static void main(String[] args) {
        // 4
        System.out.println((new Solution2()).maximalSquare(new char[][]{{'1', '0', '1', '0', '0'}, {'1', '0', '1', '1', '1'}, {'1', '1', '1', '1', '1'}, {'1', '0', '0', '1', '0'}}));
        //9
        System.out.println((new Solution2()).maximalSquare(new char[][]{{'0','0','0','1'}, {'1','1','0','1'}, {'1','1','1','1'}, {'0','1','1','1'}, {'0','1','1','1'}}));
    }
}