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 ‘.’.
A partially filled sudoku which is valid.
Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.
分析
这是一道关于数独游戏的题目，首先要了解数独游戏的规则： 
所以，对于该题目，有些空格中是’.’ 字符，我们只需要考虑当前状态下是否满足数独即可。
也就是说，我们要按行、按列，按每个3*3宫格，检验三次。
AC代码
class Solution {
public:
bool isValidSudoku(vector<vector<char>>& board) {
if (board.empty())
return false;
//数独游戏符合9宫格，也就是为一个9*9的矩阵
int size = board.size();
//根据数独游戏的规则，需要进行三次检验（按行、按列、按照3*3块）
//利用哈希的思想，记录每个关键字的出现次数，若是次数>1，则返回false
vector<int> count;
for (int i = 0; i < size; ++i)
{
//每行开始前，将记录次数的vector清零,元素1~9分别对应下标0~8,对应vector中值为该元素的出现次数
count.assign(9, 0);
for (int j = 0; j < size; j++)
{
if (board[i][j] != '.')
{
int pos = board[i][j] - '1';
if (count[pos] > 0)
return false;
else
++count[pos];
}
else
continue;
}//for
}//for
//同理，按列检验
for (int j = 0; j < size; j++)
{
count.assign(9, 0);
for (int i = 0; i < size; i++)
{
if (board[i][j] != '.')
{
int pos = board[i][j] - '1';
if (count[pos] > 0)
return false;
else
++count[pos];;
}
else
continue;
}//for
}//for
//按3*3小块检验
for (int i = 0; i < size; i += 3)
{
for (int j = 0; j < size; j += 3)
{
count.assign(9, 0);
//每个块又是一个3*3的矩阵
for (int row = i; row < i + 3;row++)
for (int col = j; col < j + 3; col++)
{
if (board[row][col] != '.')
{
int pos = board[row][col] - '1';
if (count[pos] > 0)
return false;
else
++count[pos];;
}
else
continue;
}
}//for
}//for
return true;
}
};