题目

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：
[困难] 37. 解数独 - 图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：

[困难] 37. 解数独 - 图2

解答 & 代码

递归回溯：对每个格子穷举 1 ~ 9 ，如果合法则填入，再递归处理下一格。如果找到一个可行解，则直接返回，停止搜索

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
class Solution {
private:
    // 判断数独 board[row][col] 位置能否填入数字 ch
    bool isValid(vector<vector<char>>& board, int row, int col, char ch)
    {
        for(int i = 0; i < 9; ++i)
        {
            // 检查行是否存在重复数字
            if(board[row][i] == ch)
                return false;
            // 检查列是否存在重复数字
            if(board[i][col] == ch)
                return false;
            // 检查 3*3 的宫内是否存在重复数字
            if(board[row / 3 * 3 + i / 3][col / 3 * 3 + i % 3] == ch)
                return false;
        }
        return true;
    }

    // 递归回溯
    bool backTrace(vector<vector<char>>& board, int row, int col)
    {
        // 递归结束条件：如果当前填完了所有格子，即找到了一个可行解，则直接返回 true
        if(row == 8 && col == 9)
            return true;

        // 如果当前走到行尾，则递归从下一行行首继续处理
        if(col == 9)
            return backTrace(board, row + 1, 0);
        // 如果当前的格子不是空格，已经填入数字，则跳过处理下一个
        if(board[row][col] != '.')
            return backTrace(board, row, col + 1);

        // 对当前格子，穷举 1 ~ 9 填入
        for(char ch = '1'; ch <= '9'; ++ch)
        {
            // 如果填入当前数字是合法的
            if(isValid(board, row, col, ch))
            {
                // 选择：填入当前数字
                board[row][col] = ch;
                // 递归回溯处理下一个格子，如果找到一个可行解，直接结束搜索，返回 true
                if(backTrace(board, row, col + 1) == true)
                    return true;
                // 撤销：将当前格子回复空白
                board[row][col] = '.';
            }
        }

        // 当前格子填入 1 ~ 9 都不合法，则直接返回 false
        return false;
    }
public:
    void solveSudoku(vector<vector<char>>& board) {
        backTrace(board, 0, 0);
    }
};

复杂度分析：设数独中空白格数量为 M

时间复杂度：最坏情况下，每个空白格都要枚举 9 个数，因此时间复杂度。但实际上存在剪枝，如果一个格子填入一个数不合法就不会填入，如果找到了一个可行解就会直接返回不会继续搜索
空间复杂度 O(log 81)：递归栈深度

执行结果：

1
2
3
4
执行结果：通过

执行用时：24 ms, 在所有 C++ 提交中击败了 46.20% 的用户
内存消耗：6.3 MB, 在所有 C++ 提交中击败了 55.84% 的用户

类似题目

[困难] 51. N 皇后