LeetCode 36. 有效的数独（C++）

思路：
该题就是通过三个不同的数组来存储每个位置的元素状态来判断是否有同样的元素出现，难点是将坐标转化为相应的box坐标。
原题链接：https://leetcode.cn/problems/valid-sudoku/?favorite=2ckc81c

1.题目如下：

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 ‘.’

2.代码如下：

class Solution {
public:

    bool isValidSudoku(vector<vector<char>>& board) {
        // 存储每一行的每个数是否出现过，默认初始情况下，每一行每一个数都没有出现过
        int row[9][10] = {0};
        // 整个board有9行，第二维的维数10是为了让下标有9，和数独中的数字1-9对应。
        // 存储每一列的每个数是否出现过
        int col[9][10] = {0};
        // 存储每一个box的每个数是否出现过
        int box[9][10] = {0};
        for(int i=0; i<9; i++){
            for(int j = 0; j<9; j++){
                // 遍历到第i行第j列的那个数,我们要判断这个数在其所在的行有没有出现过，
                // 同时判断这个数在其所在的列有没有出现过
                // 同时判断这个数在其所在的box中有没有出现过
                if(board[i][j] == '.') continue;
                int curNumber = board[i][j]-'0';
                if(row[i][curNumber]){
                    return false; 
                }
                if(col[j][curNumber]){
                    return false; 
                }
                //这里的重点是怎样把坐标i j转化为相应的不同的box
                if(box[j/3 + (i/3)*3][curNumber]){
                    return false; 
                }
                // 之前都没出现过，现在出现了，就给它置为1
                row[i][curNumber] = 1;
                col[j][curNumber] = 1;
                box[j/3 + (i/3)*3][curNumber] = 1;
            }
        }
        return true;
    }
};