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：

 思路一：回溯算法

bool line[9][9];
bool column[9][9];
bool block[3][3][9];
bool valid;
int* spaces[81];
int spacesSize;

void dfs(char** board, int pos) {
    if (pos == spacesSize) {
        valid = true;
        return;
    }

    int i = spaces[pos][0], j = spaces[pos][1];
    for (int digit = 0; digit < 9 && !valid; ++digit) {
        if (!line[i][digit] && !column[j][digit] && !block[i / 3][j / 3][digit]) {
            line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = true;
            board[i][j] = digit + '0' + 1;
            dfs(board, pos + 1);
            line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = false;
        }
    }
}

void solveSudoku(char** board, int boardSize, int* boardColSize) {
    memset(line, false, sizeof(line));
    memset(column, false, sizeof(column));
    memset(block, false, sizeof(block));
    valid = false;
    spacesSize = 0;

    for (int i = 0; i < 9; ++i) {
        for (int j = 0; j < 9; ++j) {
            if (board[i][j] == '.') {
                spaces[spacesSize] = malloc(sizeof(int) * 2);
                spaces[spacesSize][0] = i;
                spaces[spacesSize++][1] = j;
            } else {
                int digit = board[i][j] - '0' - 1;
                line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = true;
            }
        }
    }

    dfs(board, 0);
}

分析:

先用数组记录每个数字是否出现，对整个数独数组进行遍历，当遍历到第 i 行第 j 列的位置：如果该位置是一个空白格，那么将其加入一个用来存储空白格位置的列表中，方便后续的递归操作；如果该位置是一个数字 将 line[i][x−1]，column[j][x−1] 以及 block[⌊i/3⌋][⌊j/3⌋][x−1] 均置为 True，再使用递归枚举即可得到答案。

总结：

本题可使用回溯算法进行递归枚举，将数独补充完整后判断是否有重复格。实际考察了对递归的应用。