leetcode:37.解数独

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：

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

解答

我们可以考虑按照「行优先」的顺序依次枚举每一个空白格中填的数字，通过递归 + 回溯的方法枚举所有可能的填法。当递归到最后一个空白格后，如果仍然没有冲突，说明我们找到了答案；在递归的过程中，如果当前的空白格不能填下任何一个数字，那么就进行回溯。

由于每个数字在同一行、同一列、同一个九宫格中只会出现一次，因此我们可以使用line[i]，column[j]，block[x][y]分别表示第 i 行，第 j 列，第(x,y) 个九宫格中填写数字的情况。在下面给出的三种方法中，我们将会介绍两种不同的表示填写数字情况的方法。

九宫格的范围为 $0 \leq x \leq 2$以及 $0 \leq y \leq 2$。具体地，第 i 行第 j 列的格子位于第$(\lfloor i/3 \rfloor, \lfloor j/3 \rfloor)$个九宫格中，其中$\lfloor u \rfloor$表示对 u 向下取整。

回溯法

思路：最容易想到的方法是用一个数组记录每个数字是否出现。由于我们可以填写的数字范围为[1,9]，而数组的下标从 0 开始，因此在存储时，我们使用一个长度为 9 的布尔类型的数组，其中 i 个元素的值为 $\text{True}$，当且仅当数字i+1 出现过。例如我们用 $\textit{line}[2][3] = \text{True}$ 表示数字 4 在第 2 行已经出现过，那么当我们在遍历到第 2 行的空白格时，就不能填入数字 4。

算法：

首先对整个数独数组进行遍历，当我们遍历到第i行第j列时：

若该位置是一个空白格，那么我们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；
若该位置是一个数字x，那么我们需要将line[i][x-1],column[j][x-1]以及block[i/3][j/3][x-1]都置为True。

当我们结束了遍历过程之后，就可以开始递归枚举。当递归到第 $i$ 行第 $j$ 列的位置时，我们枚举填入的数字 $x$。根据题目的要求，数字 $x$ 不能和当前行、列、九宫格中已经填入的数字相同，因此 $line[i][x-1]$，$\textit{column}[j][x-1]$以及 $\textit{block}[\lfloor i/3 \rfloor][\lfloor j/3 \rfloor][x-1]$ 必须均为 $\text{False}$。

当我们填入了数字 $x$ 之后，我们要将上述的三个值都置为 $\text{True}$，并且继续对下一个空白格位置进行递归。在回溯到当前递归层时，我们还要将上述的三个值重新置为 $\text{False}$。

class Solution {
private:
    bool line[9][9];
    bool column[9][9];
    bool block[3][3][9];
    bool valid;
    vector<pair<int, int>> spaces;

public:
    void dfs(vector<vector<char>>& board, int pos) {
        if (pos == spaces.size()) {
            valid = true;
            return;
        }

        auto [i, j] = spaces[pos]; //获取空位位置
        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; //填符合条件的数字
                board[i][j] = digit + '0' + 1;
                dfs(board, pos + 1); //递归
                line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = false; //回溯
            }
        }
    }

    void solveSudoku(vector<vector<char>>& board) {
        memset(line, false, sizeof(line));
        memset(column, false, sizeof(column));
        memset(block, false, sizeof(block));
        valid = false;

        for (int i = 0; i < 9; ++i) {
            for (int j = 0; j < 9; ++j) {
                if (board[i][j] == '.') {
                    spaces.emplace_back(i, j); //initialize
                }
                else {
                    int digit = board[i][j] - '0' - 1;
                    line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = true; //initialize
                }
            }
        }

        dfs(board, 0); //从头开始求解，最后得到填满数字的board
    }
};

参考

力扣官方题解：https://leetcode.cn/problems/sudoku-solver/solutions/414120/jie-shu-du-by-leetcode-solution/