来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/valid-sudoku

题目描述

请你判断一个 9 x 9 的数独是否有效。只需要 根据以下规则 ，验证已经填入的数字是否有效即可。

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）

注意：

一个有效的数独（部分已被填充）不一定是可解的。

只需要根据以上规则，验证已经填入的数字是否有效即可。

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：true

示例 2：

输入：board =

[["8","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：false

解释：除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

提示：

board.length == 9

board[i].length == 9

board[i][j] 是一位数字（1-9）或者 '.'

解题思路

本题仅需要确定已经填入的数字是否符合规则，不需要判断数独是否有解，很大的降低了解题难度。

利用状态压缩的策略，每行的情况可以使用一个int来记录，其中0-8位分别代表该行1-9是否出现过，如果出现则直接可以判定这个数独不符合规则，同样每列和每个格子都可以使用这种思路。

代码展示

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        vector<int> Row(9, 0), Col(9, 0);
        vector<vector<int>> Block(3, vector<int>(3, 0));
        for(int i = 0; i < 9; i++)
            for(int j = 0; j < 9; j++)
            {
                if(board[i][j] != '.')
                {
                    int c = board[i][j] - '1';
                    if(Row[i] & 1 << c)
                        return false;
                    if(Col[j] & 1 << c)
                        return false;
                    if(Block[i / 3][j / 3] & 1 << c)
                        return false;
                    Row[i] |= 1 << c;
                    Col[j] |= 1 << c;
                    Block[i / 3][j / 3] |= 1 << c;
                }
            }
        return true;
    }
};

运行结果