題目：

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.

解法：

數獨的規則就是橫著是1-9，豎著也是1-9，每個子柵格中還是1-9，那麼我們要檢查一個沒填滿的數獨是否合法（注意不是可解），只需要檢查每一行，每一列以及每個子柵格就行了，用HashSet儲存已經遍歷過的元素。

import java.util.HashSet;


public class No35_ValidSudoku {
	public static void main(String[] args){
		
	}
	public static boolean isValidSudoku(char[][] board) {
        for(int i=0;i<board.length;i++){
        	HashSet<Character> set = new HashSet<Character>();
        	for(int j=0;j<board[i].length;j++){
        		if(board[i][j] == '.')
        			continue;
        		if(set.contains(board[i][j]))
        			return false;
        		set.add(board[i][j]);
        	}
        }
        for(int j=0;j<board[0].length;j++){
        	HashSet<Character> set = new HashSet<Character>();
        	for(int i=0;i<board.length;i++){
        		if(board[i][j] == '.')
        			continue;
        		if(set.contains(board[i][j]))
        			return false;
        		set.add(board[i][j]);
        	}
        }
        for(int i=0;i<board.length;i+=3){
        	for(int j=0;j<board[i].length;j+=3){
        		HashSet<Character> set = new HashSet<Character>();
        		for(int p=i;p<i+3;p++){
        			for(int q=j;q<j+3;q++){
        				if(board[p][q] == '.')
                			continue;
                		if(set.contains(board[p][q]))
                			return false;
                		set.add(board[p][q]);
        			}
        		}
        	}
        }
        return true;
    }
}