[LeetCode] 37. Sudoku Solver 求解数独

Write a program to solve a Sudoku puzzle by filling the empty cells.

A sudoku solution must satisfy all of the following rules:

Each of the digits 1-9 must occur exactly once in each row.
Each of the digits 1-9 must occur exactly once in each column.
Each of the the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid.

Empty cells are indicated by the character '.'.

[LeetCode] 37. Sudoku Solver 求解数独
A sudoku puzzle...

[LeetCode] 37. Sudoku Solver 求解数独
...and its solution numbers marked in red.

Note:

The given board contain only digits 1-9and the character '.'.
You may assume that the given Sudoku puzzle will have a single unique solution.
The given board size is always 9x9.

这道求解数独的题是在之前那道 Valid Sudoku 非常的相似，但更简单一些，因为这里只需要检测新加入的这个数字是否会跟其他位置引起冲突，分别检测新加入数字的行列和所在的小区间内是否有重复的数字即可，参见代码如下：

解法一：

class Solution {
public:
    void solveSudoku(vector<vector<char>>& board) {
        helper(board, 0, 0);
    }
    bool helper(vector<vector<char>>& board, int i, int j) {
        if (i == 9) return true;
        if (j >= 9) return helper(board, i + 1, 0);
        if (board[i][j] != '.') return helper(board, i, j + 1);
        for (char c = '1'; c <= '9'; ++c) {
            if (!isValid(board, i , j, c)) continue;
            board[i][j] = c;
            if (helper(board, i, j + 1)) return true;
            board[i][j] = '.';
        }
        return false;
    }
    bool isValid(vector<vector<char>>& board, int i, int j, char val) {
        for (int x = 0; x < 9; ++x) {
            if (board[x][j] == val) return false;
        }
        for (int y = 0; y < 9; ++y) {
            if (board[i][y] == val) return false;
        }
        int row = i - i % 3, col = j - j % 3;
        for (int x = 0; x < 3; ++x) {
            for (int y = 0; y < 3; ++y) {
                if (board[x + row][y + col] == val) return false;
            }
        }
        return true;
    }
};

还有另一种递归的写法，这里就不带横纵坐标参数进去，由于递归需要有 boolean 型的返回值，所以不能使用原函数。因为没有横纵坐标，所以每次遍历都需要从开头0的位置开始，这样无形中就有了大量的重复检测，导致这种解法虽然写法简洁一些，但击败率是没有上面的解法高的。这里的检测数组冲突的子函数写法也比上面简洁不少，只用了一个 for 循环，用来同时检测行列和小区间是否有冲突，注意正确的坐标转换即可，参见代码如下：

解法二：

class Solution {
public:
    void solveSudoku(vector<vector<char>>& board) {
        helper(board);
    }
    bool helper(vector<vector<char>>& board) {
        for (int i = 0; i < 9; ++i) {
            for (int j = 0; j < 9; ++j) {
                if (board[i][j] != '.') continue;
                for (char c = '1'; c <= '9'; ++c) {
                    if (!isValid(board, i, j, c)) continue;
                    board[i][j] = c;
                    if (helper(board)) return true;
                    board[i][j] = '.';
                }
                return false;
            }
        }
        return true;
    }
    bool isValid(vector<vector<char>>& board, int i, int j, char val) {
        for (int k = 0; k < 9; ++k) {
            if (board[k][j] != '.' && board[k][j] == val) return false;
            if (board[i][k] != '.' && board[i][k] == val) return false;
            int row = i / 3 * 3 + k / 3, col = j / 3 * 3 + k % 3;
            if (board[row][col] != '.' && board[row][col] == val) return false;
        }
        return true;
    }
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/37

类似题目：

Valid Sudoku

Unique Paths III

参考资料：

https://leetcode.com/problems/sudoku-solver/

https://leetcode.com/problems/sudoku-solver/discuss/15853/Simple-and-Clean-Solution-C%2B%2B

https://leetcode.com/problems/sudoku-solver/discuss/15752/Straight-Forward-Java-Solution-Using-Backtracking

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