Leetcode: Maximal Square

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

dp问题，用一个dp[i][j]保存matrix[i][j]作为右下节点的时候的最大矩形的边长

if (matrix[i][j] == '0') dp[i][j] = 0;

else dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1;

public class Solution {
    public int maximalSquare(char[][] matrix) {
        int res = 0;
        if (matrix==null || matrix.length==0 || matrix[0].length==0) return res;
        int[][] dp = new int[matrix.length][matrix[0].length];
        int maxEdge = 0;
        for (int i=0; i<matrix.length; i++) {
            if (matrix[i][0] == '1') dp[i][0] = 1;
            else dp[i][0] = 0;
            maxEdge = Math.max(maxEdge, dp[i][0]);
        }
        for (int j=1; j<matrix[0].length; j++) {
            if (matrix[0][j] == '1') dp[0][j] = 1;
            else dp[0][j] = 0;
            maxEdge = Math.max(maxEdge, dp[0][j]);
        }
        for (int i=1; i<matrix.length; i++) {
            for (int j=1; j<matrix[0].length; j++) {
                if (matrix[i][j] == '0') {
                    dp[i][j] = 0;
                    continue;
                }
                dp[i][j] = Math.min(dp[i-1][j-1], Math.min(dp[i-1][j], dp[i][j-1])) + 1;
                maxEdge = Math.max(maxEdge, dp[i][j]);
            }
        }
        return maxEdge * maxEdge;
    }
}