题目: Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.
The Sudoku board could be partially filled, where empty cells are filled with the character ‘.’.
Sudoku规则:
在一个9*9的区域内, 每行1-9出现且只出现一次, 每列1-9出现且只出现一次, 在9个子3*3的区域内1-9出现且只出现一次。
C++参考代码:
class Solution
{
public:
bool isValidSudoku(vector<vector<char> > &board)
{
//row和column分别记录行和列中元素出现的次数
int row[9], column[9];
//判断行和列是否满足要求
for (int i = 0; i < 9; i++)
{
//将row和column赋值为0
memset(row, 0, sizeof(int) * 9);
memset(column, 0, sizeof(int) * 9);
for (int j = 0; j < 9; j++)
{
//对行的判断
if (board[i][j] != '.')
{
//board[i][j] - '1'是计算board[i][j]这个数在row中对应的下标
//如果board[i][j]这个数字出现的次数大于1次,即row[board[i][j] - '1'] > 0则返回false
if (row[board[i][j] - '1'] > 0) return false;
else row[board[i][j] - '1']++;
}
//对列的判断(同理)
if (board[j][i] != '.')
{
if (column[board[j][i] - '1'] > 0) return false;
else column[board[j][i] - '1']++;
}
}
}
//判断3*3的区域是否满足要求
int block[9];
//外层的两个for循环控制对block的移动遍历
for (int i = 0; i < 9; i += 3)
{
for (int j = 0; j < 9; j += 3)
{
memset(block, 0, sizeof(int) * 9);
//内层的两个for循环控制对block内部的数据进行遍历判断
for (int m = 0; m < 3; m++)
{
for (int n = 0; n < 3; n++)
{
//判断方法和对行列的判断方法一样
if (board[i + m][j + n] != '.')
{
if (block[board[i + m][j + n] - '1'] > 0) return false;
else block[board[i + m][j + n] - '1']++;
}
}
}
}
}
return true;
}
};