leetcode--Valid Sudoku

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

A partially filled sudoku which is valid.

Here is a brute force method. I do not whether there is an other smart method.

public class Solution{
    public boolean isValidSudoku(char[][] board) {
        boolean isValid = true;
        HashSet<Character> column = new HashSet<Character>();
        HashSet<Character> row = new HashSet<Character>();
        for(int i = 0; i < 9; ++i){
            //clear the set, before we insert into a new row or column
            column.clear();
            row.clear();
            for(int j = 0; j < 9; ++j){
                if(board[i][j] != '.')
                    isValid = isValid && row.add(board[i][j]);
                if(board[j][i] != '.')
                    isValid = isValid && column.add(board[j][i]); 
                if(!isValid)
                    break;
            }
            if(!isValid)
                break;
        }
        if(isValid){
            HashSet<Character> square = new HashSet<Character>();
            int i = 0;
            while(i < 9){
                int j = 0;
                while(j < 9){
                    square.clear();
                    for(int ii = 0; ii < 3; ++ii){
                        for(int jj = 0; jj < 3; ++jj){
                            if(board[i +ii][j + jj] != '.')
                                isValid = square.add(board[i +ii][j + jj]); 
                            if(!isValid)
                                return false;
                        }
                    }
                    j += 3;
                }
                i += 3;
            }
        }
        return isValid;
    }
}