每日OJ题_DFS爆搜深搜回溯剪枝⑤_力扣37. 解数独

[力扣37. 解数独](#力扣37. 解数独)

解析代码

力扣37. 解数独

37. 解数独

难度困难

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：

复制代码

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

cpp 复制代码

class Solution {
public:
    void solveSudoku(vector<vector<char>>& board) {

    }
};

解析代码

为了存储每个位置的元素，我们需要定义⼀个⼆维数组。首先记录所有已知的数据，然后遍历所有需要处理的位置，并遍历数字 1~9。对于每个位置，检查该数字是否可以存放在该位置，同时检查行、列和九宫格是否唯一。

根据力扣36. 有效的数独 的解法，可以使用一个二维数组来记录每个数字在每一行中是否出现，一个二维数组来记录每个数字在每一列中是否出现。对于九宫格，可以用行和列除以 3 得到的商作为九宫格的坐标，并使用一个三维数组来记录每个数字在每一个九宫格中是否出现。在检查是否存在冲突时，只需检查行、列和九宫格里对应的数字是否已被标记。如果数字至少有一个位置（行、列、九宫格）被标记，则存在冲突，因此不能在该位置放置当前数字。

特别地，在本题中，我们需要直接修改给出的数组，因此在找到一种可行的方法时，应该停止递归，以防止正确的方法被覆盖。

初始化定义：

定义行、列、九宫格标记数组以及找到可行方法的标记变量，将它们初始化为 false。
定义一个数组来存储每个需要处理的位置。
将题目给出的所有元素的行、列以及九宫格坐标标记为 true。
将所有需要处理的位置存入数组。

递归流程如下：

结束条件：已经处理完所有需要处理的元素。如果找到了可行的解决方案，则将标记变量更新为true 并返回。
获取当前需要处理的元素的行列值。
遍历数字 1~9。如果当前数字可以填入当前位置，并且标记变量未被赋值为 true，则将当前位置的行、列以及九宫格坐标标记为 true，将当前数字赋值给 board 数组中的相应位置元素，然后对下一个位置进⾏递归。
递归结束时，撤回标记。

cpp 复制代码

class Solution {
    bool row[9][10];
    bool col[9][10];
    bool grid[3][3][10];

public:
    void solveSudoku(vector<vector<char>>& board) {
        for(int i = 0; i < 9; ++i) // 初始化
        {
            for(int j = 0; j < 9; ++j)
            {
                if(board[i][j] != '.')
                {
                    int n = board[i][j] - '0';
                    row[i][n] = col[j][n] = grid[i / 3][j / 3][n] = true;
                }
            }
        }
        dfs(board);
    }

    bool dfs(vector<vector<char>>& board)
    {
        for(int i = 0; i < 9; ++i)
        {
            for(int j = 0; j < 9; ++j)
            {
                if(board[i][j] == '.')
                {
                    for(int n = 1; n <= 9; ++n)
                    {
                        if(!row[i][n] && !col[j][n] && !grid[i / 3][j / 3][n])
                        {
                            board[i][j] = n + '0'; // 填数
                            row[i][n] = col[j][n] = grid[i / 3][j / 3][n] = true;
                            if(dfs(board) == true) // 填对了，告诉上一层填对了
                                return true;
                            board[i][j] = '.'; // 回复现场
                            row[i][n] = col[j][n] = grid[i / 3][j / 3][n] = false;
                        }
                    }
                    return false; // 1到9都不行
                }
            }
        }
        return true; // 填完了
    }
};