LeetCode : Valid Sudoku [java]

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、每個3*3正方形内包含1-9各一次且僅一次，是以思路就是判斷是否出現了重複數字，使用Map存儲已經出現的數字，判斷是否重複了。

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        if(board == null || board.length != 9 || board[0].length != 9){
            return false;
        }
		Map<Character, Integer> map_a = new HashMap<Character, Integer>();
		Map<Character, Integer> map_b = new HashMap<Character, Integer>();
		for (int i = 0; i < board.length; i++) {
			map_a.clear();
			map_b.clear();
			for (int j = 0; j < board[0].length; j++) {
				char ch_a = board[i][j];
				char ch_b = board[j][i];
				if (ch_a > '0' && ch_a <= '9') {
					if (map_a.containsKey(ch_a)) {
						return false;
					}
					map_a.put(ch_a,1);
				} else if (ch_a != '.') {
					return false;
				}
				if (ch_b > '0' && ch_b <= '9') {
					if (map_b.containsKey(ch_b)) {
						return false;
					}
					map_b.put(ch_b,1);
				} else if (ch_b != '.') {
					return false;
				}
			}
		}
		for (int i = 0; i < board[0].length - 2; i = i + 3) {
			for (int j = 0; j < board[0].length - 2; j = j + 3) {
				map_a.clear();
				for (int m = i; m < i + 3; m++) {
					for (int n = j; n < j + 3; n++) {
						char ch_a = board[m][n];
						if (ch_a > '0' && ch_a <= '9') {
							if (map_a.containsKey(ch_a)) {
								return false;
							}
							map_a.put(ch_a,1);
						} else if (ch_a != '.') {
							return false;
						}
					}
				}
			}
		}
		return true;
    }
}