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。

數字1-9在每一行只能出現一次。
數字1-9在每一列只能出現一次。
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。

關鍵是怎麼判斷9個小數獨，這裡用公式(i/3)*3+j/3就可以計算出當前數所在的小數獨。

 1 class Solution {
 2 public:
 3     bool isValidSudoku(vector<vector<char>>& board) {
 4         int row[9][10] = {0};// 雜湊表儲存每一行的每個數是否出現過，預設初始情況下，每一行每一個數都沒有出現過
 5         // 整個board有9行，第二維的維數10是為了讓下標有9，和數獨中的數字9對應。
 6         int col[9][10] = {0};// 儲存每一列的每個數是否出現過，預設初始情況下，每一列的每一個數都沒有出現過
 7         int box[9][10] = {0};// 儲存每一個box的每個數是否出現過，預設初始情況下，在每個box中，每個數都沒有出現過。整個board有9個box。
 8         for(int i=0; i<9; i++){
 9             for(int j = 0; j<9; j++){
10                 // 遍歷到第i行第j列的那個數,我們要判斷這個數在其所在的行有沒有出現過，
11                 // 同時判斷這個數在其所在的列有沒有出現過
12                 // 同時判斷這個數在其所在的box中有沒有出現過
13                 if(board[i][j] == '.') continue;
14                 int curNumber = board[i][j]-'0';
15                 if(row[i][curNumber]) return false; 
16                 if(col[j][curNumber]) return false;
17                 if(box[j/3 + (i/3)*3][curNumber]) return false;
18 
19                 row[i][curNumber] = 1;// 之前都沒出現過，現在出現了，就給它置為1，下次再遇見就能夠直接返回false了。
20                 col[j][curNumber] = 1;
21                 box[j/3 + (i/3)*3][curNumber] = 1;
22             }
23         }
24         return true;
25     }
26 };
27 
28 作者：liujin-4
29 連結：https://leetcode-cn.com/problems/valid-sudoku/solution/36-jiu-an-zhao-cong-zuo-wang-you-cong-shang-wang-x/
30 來源：力扣（LeetCode）
31 著作權歸作者所有。商業轉載請聯絡作者獲得授權，非商業轉載請註明出處。

用陣列直接記錄，佔用空間也比map或者set要小。