Leetcode || Valid Sudoku

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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 ‘.’.

public class Solution {  static HashMap<Character, Integer> map = new HashMap<Character, Integer>();    public void initMap() {        for(char i='0'; i<='9'; i++)                    map.put(i, 0);    }    public boolean isValidSudoku(char[][] board) {        initMap();        //行        for(int i=0; i<9; i++) {               for(int j=0; j<9; j++) {                if(board[i][j]>='0' && board[i][j]<='9') {                    if(map.get(board[i][j]) > 0)  //说明出现过                        return false;                    else                         map.put(board[i][j], 1);  //没出现过置1                }                else if(board[i][j] != '.')   //非法符号                    return false;            }        }        //列        for(int i=0; i<9; i++) {            for(int j=0; j<9; j++) {                initMap();                if(board[j][i]>='0' && board[j][i]<='9') {                    if(map.get(board[j][i]) > 0)  //说明出现过                        return false;                    else                         map.put(board[j][i], 1);  //没出现过置1                }                else if(board[j][i] != '.')   //非法符号                    return false;            }        }        //九宫        for(int i=0; i<9; i=i+3) {            for(int j=0; j<9; j=j+3) {                initMap();                for(int m=i; m<i+3; m++) {                    for(int n=j; n<j+3; n++) {                        if(board[m][n]>='0' && board[m][n]<='9') {                            if(map.get(board[m][n]) > 0)  //说明出现过                                return false;                            else                                 map.put(board[m][n], 1);  //没出现过置1                        } else if(board[m][n] != '.')   //非法符号                            return false;                    }                }            }        }        return true;    }}

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