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37. 解数独

力扣链接(困难)：https://leetcode.cn/problems/sudoku-solver

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：

  1. 数字 1-9 在每一行只能出现一次。
  2. 数字 1-9 在每一列只能出现一次。
  3. 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：

img

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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

个人题解

我有必要解释一下递归过程，首先把 9x9 大格子拆分为两个 for 循环，外层循环遍历行，内层循环遍历列。这样就分解为 81 个常规的 backtracking 模板。

C++
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for(char val = '1'; val <= '9'; val++) {
    if(!isValid(i, j, val, board)) continue;
    board[i][j] = val;
    if(backtracking(board)) return true;
    board[i][j] = '.';
}

这里的 backtracking 解决的是这样一个过程：

从 1-9 依次选数，如果合理，那就填入这个数，在当前位置填入该数字的基础上，然后开始递归。递归过程就是重复遍历这个 9x9 大格子中其它未填数字的区域，直到所有 backtracking 返回 true 这代表找到了答案。如果在某个 backtracking 中，9 个数字都不满足，代表该递归上一层所填的数字不合理，返回 false 后，上一层的 backtracking 会将之前填写的数字清除掉 board[i][j] = '.'，继续进入下一次 for(char val = '1'; val <= '9'; val++) 尝试下一个数字。

C++
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class Solution {
public:
    bool isValid(int row, int col, int val, vector<vector<char>> & board) {
        // 同一行
        for(int i = 0; i < 9; i++) {
            if(board[row][i] == val) return false;
        }
        // 同一列
        for(int i = 0; i < 9; i++) {
            if(board[i][col] == val) return false;
        }
        // 3x3
        int beginRow = row / 3 * 3;
        int beginCol = col / 3 * 3;
        for(int i = beginRow; i < beginRow + 3; i++) {
            for(int j = beginCol; j < beginCol + 3; j++) {
                if(board[i][j] == val) return false;
            }
        }

        return true;
    }
    bool backtracking(vector<vector<char>> & board) {
        // 外层循环遍历行，内层循环遍历列
        for(int i = 0; i < 9; i++) {
            for(int j = 0; j < 9; j++) {
                if(board[i][j] != '.') continue;
                for(char val = '1'; val <= '9'; val++) {
                    if(!isValid(i, j, val, board)) continue;
                    board[i][j] = val;
                    if(backtracking(board)) return true;
                    // 当前 val 不能让递归过程中所有 backtracking 返回 true
                    // 回退这个 val，寻找下一个合理数字
                    board[i][j] = '.';
                }
                // 所有数字都不满足，返回 false，让上一层更改 val
                return false;
            }
        }
        return true;
    }
    void solveSudoku(vector<vector<char>>& board) {
        backtracking(board);
    }
};