Java [leetcode 36]Valid Sudoku

题目描述：

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.

解题思路：

这道题的解题思路比较简单，因为只需要判断已经填入的数字是否合法，那么需要判断每行每列以及每个9宫格的数字是否有重复，很自然的联想到HashSet,因为每个Set里面的元素是不能相同的。顺着这个思路，那么我们为行、列及九宫格各设置9个Hashset元素的数组，每次进行判断即可。

代码如下：

public class Solution {
    public boolean isValidSudoku(char[][] board) {
		HashSet[] row = new HashSet[9];
		HashSet[] col = new HashSet[9];
		HashSet[] cell = new HashSet[9];
		for (int i = 0; i < 9; i++) {
			row[i] = new HashSet<Character>();
			col[i] = new HashSet<Character>();
			cell[i] = new HashSet<Character>();
		}
		for (int i = 0; i < 9; i++) {
			for (int j = 0; j < 9; j++) {
				if (board[i][j] != '.') {
					if (row[i].contains(board[i][j])
							|| col[j].contains(board[i][j])
							|| cell[3 * (i / 3) + j / 3].contains(board[i][j]))
						return false;
					else {
						row[i].add(board[i][j]);
						col[j].add(board[i][j]);
						cell[3 * (i / 3) + j / 3].add(board[i][j]);
					}
				}
			}
		}
		return true;
	}
}