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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：

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 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：

提示：

• board.length == 9
• board[i].length == 9
• board[i][j] 是一位数字或者 '.'
• 题目数据 保证 输入数独仅有一个

class Solution {
public:bool isvaild(int row,int col,char val,vector<vector<char>>& board){//rowfor(int i = 0;i < 9;i++){if(board[row][i] == val) return false;}//colfor(int j = 0;j < 9;j++){if(board[j][col] == val) return false;}//九宫格int startx = (row/3)*3; // 假如在第一个九宫格，row/3=0,再*3=0;int starty = (col/3)*3; //假如在第二个九宫格，row/3=1,再*3=3; 我直呼nbfor(int i = startx;i < startx+3;i++){for(int j = starty;j < starty+3;j++){if(board[i][j] == val) return false;}}return true;}bool backtracking(vector<vector<char>>& board){for(int i = 0;i < board.size();i++){for(int j = 0;j < board[0].size();j++){//遇到空格if(board[i][j] == '.'){for(char a = '1';a <= '9';a++){//判断这里应该填入啥数字合法if(isvaild(i,j,a,board)){board[i][j] = a;//得将这个状态一直返回if(backtracking(board) == true) return true;board[i][j] = '.'; // 回溯}}return false; //填入0-9都不对，都不合法，填错了。}}}return true; //填完且填正确了。}void solveSudoku(vector<vector<char>>& board) {backtracking(board);}
};