判斷一個 9x9 的數獨是否有效。只需要根據以下規則，驗證已經填入的數字是否有效即可。

1. 數字 1-9 在每一行只能出現一次。
2. 數字 1-9 在每一列只能出現一次。
3. 數字 1-9 在每一個以粗實線分隔的 3x3 宮內只能出現一次。
   
   上圖是一個部分填充的有效的數獨。

數獨部分空格內已填入了數字，空白格用 ‘.’ 表示。
示例 1:

輸入:
[
  ["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"]
]
輸出: true
 

示例 2:

輸入:
[
  ["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"]
]
輸出: false
解釋: 除了第一行的第一個數字從 5 改為 8 以外，空格內其他數字均與 示例1 相同。
     但由於位於左上角的 3x3 宮內有兩個 8 存在, 因此這個數獨是無效的。
 

說明:

• 一個有效的數獨（部分已被填充）不一定是可解的。
• 只需要根據以上規則，驗證已經填入的數字是否有效即可。
• 給定數獨序列只包含數字 1-9 和字元 ‘.’ 。
• 給定數獨永遠是 9x9 形式的。

這個題就按標記檢視行、列、3*3單元格是否數字重複即可，遍歷整個表時遇到數字將對應標誌陣列的位置置為true，如果下次發現位置已經為true說明重複，返回false。行列的標誌很好置位，單元格的陣列置位可以這樣：一共九個單元格，其座標與陣列對應位置為

(0，0)(0，1)(0，2)			(0，3)(0，4)(0，5)			(0，6)(0，7)(0，8)
(1，0)(1，1)(1，2)-->第0行	(1，3)(1，4)(1，5)-->第1行	(1，6)(1，7)(1，8)-->第2行
(2，0)(2，1)(2，2)			(2，3)(2，4)(2，5)			(2，6)(2，7)(2，8)
(3，0)(3，1)(3，2)			(3，3)(3，4)(3，5)			(3，6)(3，7)(3，8)
(4，0)(4，1)(4，2)-->第3行	(4，3)(4，4)(4，5)-->第4行	(4，6)(4，7)(4，8)-->第5行
(5，0)(5，1)(5，2)			(5，3)(5，4)(5，5)			(5，6)(5，7)(5，8)
(6，0)(6，1)(6，2)			(6，3)(6，4)(6，5)			(6，6)(6，7)(6，8)
(7，0)(7，1)(7，2)-->第8行	(7，3)(7，4)(7，5)-->第7行	(7，6)(7，7)(7，8)-->第8行
(8，0)(8，1)(8，2)			(8，3)(8，4)(8，5)			(8，6)(8，7)(8，8)
 

所以其對應關係可以描述為3*(i/3)+(j/3)。

C++原始碼：

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int m = board.size();
        int n = board[0].size();
        if (m!=9 || n!=9) return false;
        vector<vector<bool>> row(m, vector<bool>(n, false));
        vector<vector<bool>> col(m, vector<bool>(n, false));
        vector<vector<bool>> cell(m, vector<bool>(n, false));
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                int t = board[i][j];
                if(t>='1' && t<='9')
                {
                    int k = t - '1';
                    if(row[i][k] || col[k][j] || cell[3*(i/3)+j/3][k])
                        return false;
                    row[i][k] = true;
                    col[k][j] = true;
                    cell[3*(i/3)+j/3][k] = true;
                }
            }
        }
        return true;
    }
};

python3原始碼：

class Solution:
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        m = len(board)
        n = len(board[0])
        if m!=9 or n!=9:
            return False
        row = [[False for col in range(n)] for row in range(m)]
        col = [[False for col in range(n)] for row in range(m)]
        cell = [[False for col in range(n)] for row in range(m)]
        for i in range(m):
            for j in range(n):
                t = ord(board[i][j])
                if t>=ord('1') and t<=ord('9'):
                    k = t - ord('1')
                    if row[i][k] or col[k][j] or cell[3*(i//3)+(j//3)][k]:
                        return False
                    row[i][k] = True
                    col[k][j] = True
                    cell[3*(i//3)+(j//3)][k] = True
        return True

這裡要注意的是python中list初始化的時候的程式碼，這種初始化是比較推薦的，如果使用[[false]*n]*m這種方式只是淺複製，會造成只要修改一個那麼每一行都會修改的問題。所以還是採用推薦的方法去初始化列表。