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.

一刷
題解：
驗(yàn)證數(shù)獨(dú)，分三次驗(yàn)證行，列以及9個(gè)3x3子塊就可以了。用set驗(yàn)證重復(fù)問題
Time Complexity - O(n2)， Space Complexity - O(1)。

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        if(board == null || board.length != 9 || board[0].length!=9) return false;
        Set<Integer> set = new HashSet<>();
        for(int i=0; i<9; i++){//row don't have duplicate
            set.clear();
            for(int j=0; j<9; j++){
                if(board[i][j] != '.' && !set.add(board[i][j] - '0'))//empty or contains duplicate one
                    return false;
            }
        }
        
        for(int j=0; j<9; j++){//col don't have duplicate
            set.clear();
            for(int i=0; i<9; i++){
                if(board[i][j] != '.' && !set.add(board[i][j] - '0'))//empty or contains duplicate one
                    return false;
            }
        }
        
        for(int i=1; i<9; i+=3){
            for(int j=1; j<9; j+=3){
                set.clear();
                for(int k=-1; k<=1; k++){
                    for(int l = -1; l<=1; l++){
                       if(board[i+k][j+l] != '.' && !set.add(board[i+k][j+l] - '0'))//empty or contains duplicate one
                        return false;  
                    }
                }
            }
        }
        
        return true;
    }
}