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.

https://oj.leetcode.com/problems/valid-sudoku/

思路：按照规则，先检查行，然后检查列，然后检查9格。

public class Solution {	public boolean isValidSudoku(char[][] board) {		int i;		for (i = 0; i < 9; i++) {			if (!isValid(board, i, i + 1, 0, 9))				return false;		}		int j;		for (j = 0; j < 9; j++) {			if (!isValid(board, 0, 9, j, j + 1))				return false;		}		for (i = 0; i < 9; i = i + 3)			for (j = 0; j < 9; j = j + 3) {				if (!isValid(board, i, i + 3, j, j + 3))					return false;			}		return true;	}	private boolean isValid(char[][] board, int iBegin, int iEnd, int jBegin,			int jEnd) {		HashSet<Character> set = new HashSet<Character>();		for (int i = iBegin; i < iEnd; i++)			for (int j = jBegin; j < jEnd; j++) {				if (set.contains(board[i][j]))					return false;				else if (board[i][j] != ‘.‘)					set.add(board[i][j]);			}		return true;	}	public static void main(String[] args) {		char[][] board = { { ‘5‘, ‘3‘, ‘.‘, ‘.‘, ‘7‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘ },				{ ‘6‘, ‘.‘, ‘.‘, ‘1‘, ‘9‘, ‘5‘, ‘.‘, ‘.‘, ‘.‘ },				{ ‘.‘, ‘9‘, ‘8‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘6‘, ‘.‘ },				{ ‘8‘, ‘.‘, ‘.‘, ‘.‘, ‘6‘, ‘.‘, ‘.‘, ‘.‘, ‘3‘ },				{ ‘4‘, ‘.‘, ‘.‘, ‘8‘, ‘.‘, ‘3‘, ‘.‘, ‘.‘, ‘1‘ },				{ ‘7‘, ‘.‘, ‘.‘, ‘.‘, ‘2‘, ‘.‘, ‘.‘, ‘.‘, ‘6‘ },				{ ‘.‘, ‘6‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘2‘, ‘8‘, ‘.‘ },				{ ‘.‘, ‘.‘, ‘.‘, ‘4‘, ‘1‘, ‘9‘, ‘.‘, ‘.‘, ‘5‘ },				{ ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘8‘, ‘.‘, ‘.‘, ‘7‘, ‘9‘ } };		System.out.println(new Solution().isValidSudoku(board));	}}