【Leetcode】Valid Sudoku

题目链接：https://leetcode.com/problems/valid-sudoku/

题目：

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.

思路：

每行、每列、每个黑粗线内小方块的数分别用HashSet保存判断是否唯一。

算法：

</pre><pre name="code" class="java">	public boolean isValidSudoku(char[][] board) {
		Set<Character> row = new HashSet<Character>();
		Set<Character> column = new HashSet<Character>();
		Set<Character> subBox = new HashSet<Character>();

		for (int i = 0; i < board.length; i++) {
			for (int j = 0; j < board[0].length; j++) {
				// System.out.print(board[i][j]+" ");
				if (board[i][j] != '.') {
					if (column.contains(board[i][j])) {
						return false;
					} else {
						column.add(board[i][j]);
					}
				}
				// System.out.println();
			}
			column.clear();
		}

		// System.out.println("=============");
		for (int i = 0; i < board.length; i++) {
			for (int j = 0; j < board[0].length; j++) {
				if (board[j][i] != '.') {
					if (row.contains(board[j][i])) {
						return false;
					} else {
						row.add(board[j][i]);
					}
				}
			}
			row.clear();
		}

		for (int i = 0; i < board.length; i += 3) {
			for (int j = 0; j < board[0].length; j += 3) {
				for (int ii = i; ii < i + 3; ii++) {
					for (int jj = j; jj < j + 3; jj++) {
						if (board[ii][jj] != '.') {
							if (subBox.contains(board[ii][jj])) {
								return false;
							} else {
								subBox.add(board[ii][jj]);
							}
						}
					}
				}
				subBox.clear();
			}
		}
		return true;
	}