LeetCode – Maximal Square (Java)
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:
1101 1101 1111
Return 4.
Analysis
This problem can be solved by dynamic programming. The changing condition is:
t[i][j] = min(t[i][j-1], t[i-1][j], t[i-1][j-1]) + 1. It means the square formed before this point.
Java Solution
public int maximalSquare(char[][] matrix) { if(matrix==null||matrix.length==0){ return 0; } int result=0; int[][] dp = new int[matrix.length][matrix[0].length]; for(int i=0; i<matrix.length; i++){ dp[i][0]=matrix[i][0]-'0'; result=Math.max(result, dp[i][0]); } for(int j=0; j<matrix[0].length; j++){ dp[0][j]=matrix[0][j]-'0'; result=Math.max(result, dp[0][j]); } for(int i=1; i<matrix.length; i++){ for(int j=1; j<matrix[0].length; j++){ if(matrix[i][j]=='1'){ int min = Math.min(dp[i-1][j], dp[i][j-1]); min = Math.min(min, dp[i-1][j-1]); dp[i][j]=min+1; result = Math.max(result, min+1); }else{ dp[i][j]=0; } } } return result*result; } |
<pre><code> String foo = "bar"; </code></pre>
-
Israel Lopez
-
Alina Rozenbaum
-
Guest