【LintCode】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.

样例
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.

解法：动态规划
参考：http://blog.youkuaiyun.com/brucehb/article/details/46384411
dp[i][j]表示以节点matrix[i][j]为正方形右下角可构成的最大正方形的边长。
当matrix[i][j]为1时
dp[i][j] = min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) + 1。
当matrix[i][j]为0时
dp[i][j] = 0.

public class Solution {
    /**
     * @param matrix: a matrix of 0 and 1
     * @return: an integer
     */
    public int maxSquare(int[][] matrix) {
        if(null == matrix || matrix.length == 0 || matrix[0].length == 0) return 0;
        int [][] dp = new int[matrix.length][matrix[0].length];
        int maxsize = 0;
        for(int i = 0; i < matrix.length; i++) {
            if(matrix[i][0] == 1){
                dp[i][0] = 1;
                maxsize = 1;
            }
        }

        for(int i = 0; i < matrix[0].length; i++) {
            if(matrix[0][i] == 1) {
                dp[0][i] = 1;
                maxsize = 1;
            }
        }

        for(int i = 1; i < matrix.length; i++) {
            for(int j = 1; j < matrix[0].length; j++) {
                if(matrix[i][j] == 1) {
                    dp[i][j] = Math.min(Math.min(dp[i-1][j-1], dp[i-1][j]), dp[i][j-1]) + 1;
                    maxsize = Math.max(maxsize, dp[i][j]);
                }                
            }
        }
        return maxsize * maxsize;
    }
}