221 Maximal Square

221. Maximal Square

1. Question

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.

2. Implementation

(1) DP

思路: 这类在board上的dp题基本都万变不离其中, (i,j)的状态基本受到其周围的cell的影响。这题里，我们定义dp[i][j]为以(i,j)作为正方形的右下角，该正方形的最长长度。显然要构成正方形必须要满足两个条件:

(1) (i, j)上的数字是1

(2) 由于我们以(i, j)作为正方形的右下角，所以(i - 1, j), (i, j - 1), (i - 1, j- 1)这三点也必须要构成正方形, 正方形的长度取决于这三个点的最小值

所以状态转移方程为 dp[i][j] = 1 + min(dp[i -1][j], dp[i][j - 1], dp[i - 1][j - 1]). 要求出最大正方形，只要找出最大的正方形边，然后计算面积即可

class Solution {
    public int maximalSquare(char[][] matrix) {
        if (matrix == null || matrix.length == 0) {
            return 0;
        }

        int m = matrix.length, n = matrix[0].length;
        // dp[i][j] means the edge of square which is at bottom right at (i, j)
        int[][] dp = new int[m][n];
        int maxSize = 0;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 || j == 0) {
                    dp[i][j] = matrix[i][j] - '0';
                }
                else if (matrix[i][j] == '1') {
                    dp[i][j] = 1 + Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1]));
                }
                else {
                    dp[i][j] = 0;
                }

                maxSize = Math.max(maxSize, dp[i][j]);
            }
        }
        return maxSize * maxSize;
    }
}

3. Time & Space Complexity

DP: 时间复杂度O(mn), 空间复杂度O(mn)