有效的数独——C

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。

思路：

首先先检测行，然后检测列，最后检测三比三小格里面的东西

无论是检测行还是列都可以用标记的方法，设立一个数组flag，如果在那一行中flag[1]出现一次，那么flag[1]就++，最后检测，如果最后对应的值是2就返回false

三比三小格是固定的，在固定范围里面用行或者列覆盖就可以了

代码：

bool isValidSudoku(char** board, int boardSize, int* boardColSize){
 int count = 0;
    int *flag = (int*)malloc(sizeof(int) * 10);
    for(int i = 0; i < 9; i++){
    for(int k = 0; k <= 9; k++)flag[k] = 0;
    for(int j = 0; j < 9; j++){
        if(board[i][j] != '.'){
            flag[board[i][j] - 48]++;
            if(flag[board[i][j] - 48] == 2)return false;
        }
    }
    }
     for(int i = 0; i < 9; i++){
        for(int k = 0; k <= 9; k++)flag[k] = 0;
        for(int j = 0; j < 9; j++){
            if(board[j][i] != '.'){
                flag[board[j][i] - 48]++;
                if(flag[board[j][i] - 48] == 2)return false;
            }
        }
     }
    for(int i = 0; i < 7; i += 3){
        for(int k = 0; k <= 9; k++)flag[k] = 0;
        for(int j = 0; j < 9; j++){
                if(count % 3 == 0)
                for(int k = 0; k <= 9; k++)flag[k] = 0;
            if(board[i][j] != '.'){
                flag[board[i][j] - 48]++;
                if(flag[board[i][j] - 48] == 2)return false;
            }
            if(board[i + 1][j] != '.'){
                flag[board[i + 1][j] - 48]++;
                if(flag[board[i + 1][j] - 48] == 2)return false;
            }
            if(board[i + 2][j] != '.'){
                flag[board[i + 2][j] - 48]++;
               if(flag[board[i + 2][j] - 48] == 2)return false;
            }        
            count++;
        }
    }
    return true;
}