问题描写叙述：

Determine if a Sudoku is valid, according to: .

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

A partially filled sudoku which is valid.

Note:

A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

代码：

public class Valid_Sudoku {  //java	public boolean isValidSudoku(char[][] board) {        		int size = 9;		int [] member = new int[size];   //record if is occur; 				//valid row		for(int i = 0; i < 9; i++){						for(int k = 0; k < 9; k++)				member[k] = 0;			for(int j = 0; j < 9; j++){				if(board[i][j] == '.')					continue;				int pos = board[i][j]-'0';				if(member[pos-1] == 1)					return false;				else member[pos-1] = 1;			}		}				//valid col		for(int i = 0; i < 9; i++){			for(int k = 0; k < 9; k++)				member[k] = 0;			for(int j = 0; j < 9; j++){				if(board[j][i] == '.')					continue;				int pos = board[j][i]-'0';				if(member[pos-1] == 1)					return false;				else member[pos-1] = 1;			}		}				//valid cube		for(int ibegin = 0; ibegin < 9; ibegin = ibegin+3){			for(int jbegin = 0; jbegin < 9; jbegin = jbegin+3){				for(int k = 0; k < 9; k++)					member[k] = 0;				for(int i = ibegin; i < ibegin+3; i++){					for(int j = jbegin; j < jbegin+3; j++){						if(board[i][j] == '.')							continue;						int pos = board[i][j]-'0';						if(member[pos-1] == 1)							return false;						else member[pos-1] = 1;					}				}			}		}		return true;    }		public static void main(String [] args){		String[] boardStr = {"......5..",							 ".........",							 ".........",							 "93..2.4..",							 "..7...3..",							 ".........",							 "...34....",							 ".....3...",							 ".....52.."};		char [][] board = new char [9][9];		for(int i =0; i< boardStr.length; i++){			for(int j = 0; j