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：
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 输入数独仅有一个解
#include<iostream>
#include<stack>
#include<algorithm>
#include<string>
#include<vector>
using namespace std;
/*
对于每个需要填数字的格子带入1到9，每代入一个数字都判定其是否合法，
如果合法就继续下一次递归，结束时把数字设回 '.'，
判断新加入的数字是否合法时，只需要判定当前数字是否合法，
不需要判定这个数组是否为数独数组，因为之前加进的数字都是合法的，
这样可以使程序更加高效一些，整体思路是这样的，
但是实现起来可以有不同的形式。
一种实现形式是递归带上横纵坐标，由于是一行一行的填数字，
且是从0行开始的，所以当i到达9的时候，
说明所有的数字都成功的填入了，直接返回 ture。
当j大于等于9时，当前行填完了，需要换到下一行继续填，则继续调用递归函数，
横坐标带入 i+1。否则看若当前数字不为点，
说明当前位置不需要填数字，则对右边的位置调用递归。
若当前位置需要填数字，则应该尝试填入1到9内的所有数字，
让c从1遍历到9，每当试着填入一个数字，
都需要检验是否有冲突，使用另一个子函数 isValid 来检验是否合法，
假如不合法，则跳过当前数字。若合法，则将当前位置赋值为这个数字，
并对右边位置调用递归，若递归函数返回 true，
则说明可以成功填充，直接返回 true。不行的话，需要重置状态，
将当前位置恢复为点。若所有数字都尝试了，还是不行，则最终返回 false。
*/
class Solution {
public:
void solveSudoku(vector<vector<char>>& board)
{
helper(board);
}
bool helper(vector<vector<char>>& board)
{
for (int i = 0; i < 9; ++i)
{
for (int j = 0; j < 9; ++j)
{
if (board[i][j] != '.')
{
continue;
}
for (char c = '1'; c <= '9'; ++c)
{
if (!isValid(board, i, j, c))
{
continue;
}
board[i][j] = c;
if (helper(board))
{
return true;
}
board[i][j] = '.';
}
return false;
}
}
return true;
}
bool isValid(vector<vector<char>>& board, int i, int j, char val)
{
for (int k = 0; k < 9; ++k)
{
if (board[k][j] != '.' && board[k][j] == val)
{
return false;
}
if (board[i][k] != '.' && board[i][k] == val)
{
return false;
}
int row = i / 3 * 3 + k / 3, col = j / 3 * 3 + k % 3;
if (board[row][col] != '.' && board[row][col] == val)
{
return false;
}
}
return true;
}
};