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、每个3*3正方形内包含1-9各一次且仅一次，因此思路就是判断是否出现了重复数字，使用Map存储已经出现的数字，判断是否重复了。

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;    }}