数据结构003：有效的数独

原文链接：数据结构003：有效的数独

题目

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。

题解

根据题目的规则，数独需要满足三个规则，针对规则一和二可知，我们在遍历每个元素的时候，需要判断该元素所在行和列中是否出现过，即可判断该元素是否满足规则一和二，因此我们可以针对每一行、每一列出现元素的次数作为校验标准，例如声明两个二维数组row[9][9] 和col[9][9] 分别代表行和列上面0-9 出现的次数。例如row[1][2] 表示第1行中，出现2的次数，col[4][3] 表示第4列出现3的次数（都是从第0行/列开始算的）。对于数独数组第i 行j 列上的数值n=board[i][j] ，首先将row[i][n] 上对应的值加一，再将col[j][n] 也加一，然后判断row[i][n] 和row[i][n] 的值是否大于1，大于1则表明i 行或者j 列数字n 出现的次数大于1，即不唯一。不满足规则一或者二。

对于规则三，我们可以根据元素board[i][j] 的i 和j 的索引除以3来进行判断其属于哪个小九宫格，即其对应的小九宫格的索引为i/3 和j/3 。因此我们可构建一个box[3][3][9] 的三位数组来记录每个小九宫格中0-9 出现的次数，例如box[1][2][3] 表示第一行第二列的九宫格中出现数字3的次数，我们的思路与row 和col 一样，遍历每个元素n=board[i][j] ，并将box[i/3][j/3][n] 的值加一，在判断其是否大于1。

通过上面的分析，我们的实现代码如下：

class Solution {
    public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int row[9][9] = {0};
        int col[9][9] = {0};
        int box[3][3][9] = {0};

        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                char c = board[i][j];
                if (c == '.') continue;
                int n = c - '1';
                row[i][n]++;
                col[j][n]++;
                box[i / 3][j / 3][n]++;
                if (row[i][n] > 1 || col[j][n] > 1 || box[i / 3][j / 3][n] > 1) {
                    return false;
                }
            }
        }
        return true;
    }
};

由于数独共有81个单元格，只需要对每个单元格遍历一次即可，因此其时间复杂度为O(1) 。由于数独的大小固定，因此空间的大小也是固定的，空间复杂度也为O(1) 。