221. Maximal Square

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.

For example, given the following matrix:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Return 4.

本题是正方形，
矩形：85. Maximal Rectangle

Solution：DP

思路：dp[i][j]存的是 以i-1, j-1作为右下点结束 的最长正方形边长
则状态转移方程为dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;
Time Complexity: O(mn) Space Complexity: O(mn)

Solution Code：

class Solution {
    public int maximalSquare(char[][] matrix) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        
        int m = matrix.length;
        int n = matrix[0].length;
        
        int[][] dp = new int[m + 1][n + 1];
        
        int max_length = 0;
        for(int i = 1; i < m + 1; i++) {
            for(int j = 1; j < n + 1; j++) {
                int row = i - 1;
                int col = j - 1;
                if(matrix[row][col] == '1') {
                    dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;
                    if(dp[i][j] > max_length) {
                        max_length = dp[i][j];
                    }
                }
            }
        }
        return max_length * max_length;
    }
}