Link: https://leetcode.cn/problems/maximal-square/

Question

difficulty: mid
adj diff: 3

Given an m x n binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.

Example 1:

Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
Output: 4

Example 2:

Input: matrix = [["0","1"],["1","0"]]
Output: 1

Example 3:

Input: matrix = [["0"]]
Output: 0

Constraints:

	m == matrix.length
	n == matrix[i].length
	1 <= m, n <= 300
	matrix[i][j] is '0' or '1'.

这道题的重点是 dp 状态转移方程 怎么写：

如下：

dp(i,j)=min(dp(i−1,j),dp(i−1,j−1),dp(i,j−1))+1

一个例子，如下。

0 1 1 1 0
1 1 1 1 0
0 1 1 1 1
0 1 1 1 1
0 0 1 1 1

对应的 dp 值如下。

0 1 1 1 0
1 1 2 2 0
0 1 2 3 1
0 1 2 3 2
0 0 1 2 3

Code

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public int maximalSquare(char[][] matrix) {
    int m = matrix.length;
    int n = matrix[0].length;
    int[][] ans = new int[m][n];
    int max = 0;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (matrix[i][j] == '0') {
                ans[i][j] = 0;
            } else if (i == 0 || j == 0) {
                ans[i][j] = 1;
            } else {
                ans[i][j] = Math.min(Math.min(ans[i-1][j], ans[i][j-1]), ans[i-1][j-1]) + 1;
            }
            max = Math.max(max, ans[i][j]);
        }
    }
    return max * max;
}