原文链接: https://leetcode-cn.com/problems/sudoku-solver

英文原文

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

A sudoku solution must satisfy all of the following rules:

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

The '.' character indicates empty cells.

 

Example 1:

Input: 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"]]
Output: [["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"]]
Explanation: The input board is shown above and the only valid solution is shown below:

 

Constraints:

• board.length == 9
• board[i].length == 9
• board[i][j] is a digit or '.'.
• It is guaranteed that the input board has only one solution.

中文题目

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：

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：

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

通过代码

高赞题解

思路

1. 状态压缩

   11. 使用 bitset<9> 来压缩存储每一行、每一列、每一个 3x3 宫格中 1-9 是否出现

   12. 这样每一个格子就可以计算出所有不能填的数字，然后得到所有能填的数字 getPossibleStatus()

   13. 填入数字和回溯时，只需要更新存储信息

   14. 每个格子在使用时，会根据存储信息重新计算能填的数字

2. 回溯

   21. 每次都使用 getNext() 选择能填的数字最少的格子开始填，这样填错的概率最小，回溯次数也会变少

   22. 使用 fillNum() 在填入和回溯时负责更新存储信息

   23. 一旦全部填写成功，一路返回 true ，结束递归

图解

答题

[]

class Solution {

public:

    bitset<9> getPossibleStatus(int x, int y)

    {

        return ~(rows[x] | cols[y] | cells[x / 3][y / 3]);

    }



    vector<int> getNext(vector<vector<char>>& board)

    {

        vector<int> ret;

        int minCnt = 10;

        for (int i = 0; i < board.size(); i++)

        {

            for (int j = 0; j < board[i].size(); j++)

            {

                if (board[i][j] != '.') continue;

                auto cur = getPossibleStatus(i, j);

                if (cur.count() >= minCnt) continue;

                ret = { i, j };

                minCnt = cur.count();

            }

        }

        return ret;

    }



    void fillNum(int x, int y, int n, bool fillFlag)

    {

        rows[x][n] = (fillFlag) ? 1 : 0;

        cols[y][n] = (fillFlag) ? 1 : 0;

        cells[x/3][y/3][n] = (fillFlag) ? 1: 0;

    }

    

    bool dfs(vector<vector<char>>& board, int cnt)

    {

        if (cnt == 0) return true;



        auto next = getNext(board);

        auto bits = getPossibleStatus(next[0], next[1]);

        for (int n = 0; n < bits.size(); n++)

        {

            if (!bits.test(n)) continue;

            fillNum(next[0], next[1], n, true);

            board[next[0]][next[1]] = n + '1';

            if (dfs(board, cnt - 1)) return true;

            board[next[0]][next[1]] = '.';

            fillNum(next[0], next[1], n, false);

        }

        return false;

    }



    void solveSudoku(vector<vector<char>>& board) 

    {

        rows = vector<bitset<9>>(9, bitset<9>());

        cols = vector<bitset<9>>(9, bitset<9>());

        cells = vector<vector<bitset<9>>>(3, vector<bitset<9>>(3, bitset<9>()));



        int cnt = 0;

        for (int i = 0; i < board.size(); i++)

        {

            for (int j = 0; j < board[i].size(); j++)

            {

                cnt += (board[i][j] == '.');

                if (board[i][j] == '.') continue;

                int n = board[i][j] - '1';

                rows[i] |= (1 << n);

                cols[j] |= (1 << n);

                cells[i / 3][j / 3] |= (1 << n);

            }

        }

        dfs(board, cnt);

    }



private:

    vector<bitset<9>> rows;

    vector<bitset<9>> cols;

    vector<vector<bitset<9>>> cells;

};

感谢 @cocowy 提供的 bitset 风格的 fillNum() ，可读性得到质的提升，已更新到上面

下面是我原来位运算风格的 fillNum()

[]

void fillNum(int x, int y, int n, bool fillFlag)

{

    bitset<9> pick(1 << n);

    rows[x] = (fillFlag) ? (rows[x] | pick) : (rows[x] ^ pick);

    cols[y] = (fillFlag) ? (cols[y] | pick) : (cols[y] ^ pick);

    cells[x / 3][y / 3] = (fillFlag) ? (cells[x / 3][y / 3] | pick) : (cells[x / 3][y / 3] ^ pick);

}

执行时间

对格子填入顺序的选择，使得搜索效率大大提升（加上运气好），击败 100%

得益于状态压缩，内存的使用也击败 100%

（别太较真儿，看个乐就好）

致谢

感谢您的观看，希望对您有帮助，欢迎热烈的交流！

如果感觉还不错就点个赞吧~

统计信息

通过次数	提交次数	AC比率
109590	163431	67.1%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言

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