[数组]36. 有效的数独-简单

​​36. 有效的数独​​

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。

来源：力扣（LeetCode）

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

著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

题解

思路：

1. 使用3个9*9的矩阵row[i][num]、col[i][num]、box[r][c]作为hash表，遍历board[i][k]，i表示行，k表示列，num表示board[i][k]所代表的数字-1
2. row[i][num]:表示第i行上是否出现过num
3. col[k][num]:表示第k列是否出现过num
4. box[r][c]:r = i/3x3+num/3，c=k/3x3+num%3，表示在3*3的矩阵中num是否出现过

代码如下：

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int row[9][9] = {0};// row[i][k]表示第i行的数字k是否出现过，为0表示没有出现过，为1表示出现过
        int col[9][9] = {0};
        int box[9][9] = {0};// 从左到右、从上到下，每3*3的9个格子组成一个大格子，在这个格子里面是否出现过数字
        for (int i = 0;i < 9;++i)
        {
            for (int k = 0;k < 9;++k)
            {
                char x = board[i][k];
                if (x == '.')
                {
                    continue;
                }

                // 在第i行、k列或者在Box中，这个数字是否出现过
                int num = x - '0' -1;
                int  r = i /3 * 3 + num / 3;
                int c = k / 3 * 3 + num % 3;
                if (row[i][num] == 1 || col[k][num] == 1 || box[r][c] == 1)
                {
                    return false;
                }
                row[i][num] = col[k][num] = box[r][c] = 1;
            }
        }
        return true;
    }
};