[leetcode]Valid Sudoku

问题描述：

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.
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<boardStr[i].length(); j++){
board[i][j] = boardStr[i].charAt(j);
}
}

Valid_Sudoku vs = new Valid_Sudoku();
System.out.println(vs.isValidSudoku(board));
}
}