Java for LeetCode 036 Valid Sudoku

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 ‘.‘.

解题思路：

传说中的数独(九宫格)问题，老实遍历三个规则即可：

JAVA实现：

	static public boolean isValidSudoku(char[][] board) {
		for (int i = 0; i < board.length; i++) {
			HashMap<Character, Integer> hashmap = new HashMap<Character, Integer>();
			for (int j = 0; j < board[0].length; j++) {
				if (board[i][j] != ‘.‘) {
					if (hashmap.containsKey(board[i][j]))
						return false;
					hashmap.put(board[i][j], 1);
				}
			}
		}
		for (int j = 0; j < board[0].length; j++) {
			HashMap<Character, Integer> hashmap = new HashMap<Character, Integer>();
			for (int i = 0; i < board.length; i++) {
				if (board[i][j] != ‘.‘) {
					if (hashmap.containsKey(board[i][j]))
						return false;
					hashmap.put(board[i][j], 1);
				}
			}
		}
	    for (int i = 0; i < board.length; i += 3){
	       for (int j = 0; j < board[0].length; j += 3){
	           HashMap<Character, Integer> hashmap = new HashMap<Character, Integer>();
				for (int k = 0; k < 9; k++) {
					if (board[i + k / 3][j + k % 3] != ‘.‘) {
						if (hashmap.containsKey(board[i + k / 3][j + k % 3]))
							return false;
						hashmap.put(board[i + k / 3][j + k % 3], 1);
					}
				}
			}
		}
		return true;
	}