Valid Sudoku

因为题目才初步了解了数独，原来一直以为跟中国的九宫格类似。既然要判断唯一性，那很自然地想到hash，用一维数组判断单列数独就够了。单行单列都好判断，难的是小3 x 3方格内的判断。思路只想到了分为3大行，后来看了解答知道还应该再分3大列。解答的那几重循环还是挺巧的，自己写估计得琢磨半天。

代码如下：

int hashSudoku(int *hashtable, int number)
{
    int index = number % 10 - 1;
    if (hashtable[index] != number) {
        hashtable[index] = number;
        return 0;
    }
    else {
        return 1;
    }
}
void resetHashTable(int *hashtable)
{
    for (int i = 0; i < 9; i++) {
        hashtable[i] = -1;
    }
}
bool isValidSudoku(char board[9][9]) {
    bool result = true;
    int hashcolumns = 9;
//    int (*hashtable)[9] = (int (*)[9])malloc(hashrows * hashcolumns * sizeof(int));//注意内存泄露
    int hashtable[9];
    for (int i = 0; i < hashcolumns; i++)
        hashtable[i] = i;
    
    for (int i = 0; i < 9; i++) {
        for (int j = 0; j < 9; j++)
            if ( board[i][j] != '.' && hashSudoku(hashtable, board[i][j] - '0') ) { //检测行
                return false;
            }
        resetHashTable(hashtable);
        
        for (int j = 0; j < 9; j++)
            if ( board[j][i] != '.' && hashSudoku(hashtable, board[j][i] - '0') ) { //检测列
                return false;
            }
        resetHashTable(hashtable);
    }
    
    for (int bigrow = 0; bigrow < 3; bigrow++) {
        for (int bigcolumn = 0; bigcolumn < 3; bigcolumn++) {
            resetHashTable(hashtable);
            for (int row = bigrow * 3; row < bigrow * 3 + 3; row++) {
                for (int column = bigcolumn * 3; column < bigcolumn * 3 + 3; column++) {
                    if ( board[row][column] != '.' && hashSudoku(hashtable, board[row][column] - '0') ) { //检测列
                        return false;
                    }
                }
            }
        }
    }
    
    return result;
}