221. Maximal Square java solutions

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.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

题目是要找出元素全为1 的最大正方形。使用DP，递推式为：

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

dp[i][j] 表示到该位置出现的正方形的边长，dp[i][j] 要构成一个新的正方形，只能从之前的左边，上边，左上 三个里面取边长最小的加上matrix[i][j] 为1 的情况，即边长+1.

 1 public class Solution {
 2     public int maximalSquare(char[][] matrix) {
 3         if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0;
 4         int m = matrix.length,n = matrix[0].length;
 5         int[][] dp = new int[m][n];
 6         int len = 0;
 7         for(int i = 0; i<m; i++){
 8             if(matrix[i][0] == ‘1‘){
 9                 dp[i][0] = 1;
10                 len = 1;
11             }
12         }
13         for(int i = 0; i<n; i++){
14             if(matrix[0][i] == ‘1‘){
15                 dp[0][i] = 1;
16                 len = 1;
17             }
18         }
19         for(int i = 1; i < m; i++){
20             for(int j = 1; j < n; j++){
21                 if(matrix[i][j] == ‘1‘){
22                     dp[i][j] = Math.min(dp[i-1][j],Math.min(dp[i][j-1],dp[i-1][j-1])) + 1;
23                     len = dp[i][j] > len ? dp[i][j] : len;
24                 }
25             }
26         }
27         return len*len;
28     }
29 }