编写一个程序，通过填充空格来解决数独问题。
数独的解法需 遵循如下规则：
- 数字
1-9在每一行只能出现一次。 - 数字
1-9在每一列只能出现一次。 - 数字
1-9在每一个以粗实线分隔的3x3宫内只能出现一次。（请参考示例图）
数独部分空格内已填入了数字，空白格用 '.' 表示。
示例 1：
输入：board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]]
输出：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：class Solution {
boolean[][] row;
boolean[][] col;
boolean[][][] map = new boolean[3][3][10];
int n;
int[][] dir = new int[][]{{1,0},{-1,0},{0,1},{0,-1}};
public void solveSudoku(char[][] board) {
n = 9;
row = new boolean[9][10];
col = new boolean[9][10];
int cnt = 0;
for(int i = 0 ; i < n ; i++){
for(int j = 0 ; j < n ; j++){
if(board[i][j] != '.'){
int t = board[i][j] - '0';
row[i][t] = true;
col[j][t] = true;
map[i/3][j/3][t] = true;
}else｛
cnt++;
｝
}
}
boolean[][] vis = new boolean[9][9];
dfs(board, 0, 0, cnt, vis);
}
boolean dfs(char[][] board , int i , int j , int cnt , boolean[][] vis){
if(cnt == 0) return true;
if(vis[i][j]) return false;
for(int[] d : dir){
int x = i + d[0];
int y = j + d[1];
if(x>=0 && y >= 0 && x < n && y < n ){
if(board[x][y] == '.'){
int target = -1;
for(int a = 1 ; a <= 9 ; a++){
if(!row[x][a] && !col[y][a] && !map[x/3][y/3][a] && !vis[x][y]){
target = a;
board[x][y] = (char)('0' + target);
row[x][target] = true;
col[y][target] = true;
map[x/3][y/3][target] = true;
vis[x][y] = true;
if(dfs(board,x,y,cnt-1,vis)) return true;
vis[x][y] = false;
row[x][target] = false;
col[y][target] = false;
map[x/3][y/3][target] = false;
board[x][y] = '.';
}
}
}else{
vis[i][j] = true;
if(dfs(board,x,y,cnt,vis)) return true;
vis[i][j] = false;
}
}
}
return false;
}
}
