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.

思路分析：這題考察DP。緩存中間結果降低反復計算。DP方程為

if(matrix[i][j] != 0) dp[i,j] = min{dp[i-1,j], dp[i,j-1], dp[i-1,j-1]} + 1; else dp[i,j] = 0

當中dp[i,j]表示以[i,j]為右下角的區域內的最大的正方形的邊長，最後返回dp數組中最大值的平方即為所求。

AC Code

public class Solution {
    public int maximalSquare(char[][] matrix) {
        //dp equation: if(matrix[i][j] != 0) dp[i,j] = min{dp[i-1,j], dp[i,j-1], dp[i-1,j-1]} + 1; else dp[i,j] = 0
        int m = matrix.length;
        if(m == 0) return 0;
        int n = matrix[0].length;
        //1227
        int [][] dp = new int[m][n];
        int max = 0;
        for(int i = 0; i < m; i++){
            if(matrix[i][0] == ‘1‘) {
                dp[i][0] = 1;
                max = Math.max(max, dp[i][0]);
            } 
        }
        
        for(int j = 0; j < n; j++){
            if(matrix[0][j] == ‘1‘) {
                dp[0][j] = 1;
                max = Math.max(max, dp[0][j]);
            } 
        }
        
        for(int i = 1; i < m; i++){
            for(int j = 1; j < n; j++){
                if(matrix[i][j] == ‘1‘){
                    dp[i][j] =  Math.min(Math.min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]) + 1;
                    max = Math.max(max, dp[i][j]);
                } else {
                    dp[i][j] = 0;
                }
            }
        }
        return max * max;
    }
    //1234
}

LeetCode Maximal Square