221. Maximal Square

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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:

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

Return 4.

解法1：

诀窍在于dp[i][j] 表示以(i,j)为底的正方形的边长。
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1, if matrix[i][j] == 1,
画一下图就比较清楚。
C++

1

Java

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public class Solution {
    public int maximalSquare(char[][] matrix) {
        if (matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        int row = matrix.length;
        int col = matrix[0].length;
        int[][] dp = new int[row][col];
        dp[0][0] = matrix[0][0] - '0' == 1 ? 1 : 0;
        int res = dp[0][0];    // record the final result
        for (int j = 1; j < col; j++) {
            if (matrix[0][j] - '0' == 1) {
                dp[0][j] = 1;
                res = 1;
            }
        }
        for (int i = 1; i < row; i++) {
            if (matrix[i][0] - '0' == 1) {
                dp[i][0] = 1;
                res = 1;
            }
        }
        for (int i = 1; i < row; i++) {
            for (int j = 1; j < col; j++) {
                if (matrix[i][j] - '0' == 0) {
                    dp[i][j] = 0;
                } else {
                    dp[i][j] = Math.min(Math.min(dp[i - 1][j], dp[i][j - 1]),dp[i - 1][j - 1]) + 1;
                }
                res = Math.max(res, dp[i][j] * dp[i][j]);
            }
        }
        return res;        
    }
}