有效的數獨

請你判斷一個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。（請參考示例圖）

數獨部分空格內已填入了數字，空白格用'.'表示。

注意：

• 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。

示例 1：

輸入：board = 
[["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

遍歷就行了：

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int book[10];
        memset(book,0,sizeof(book));
        for(int i = 0; i < 9; i++){
            for(int j = 0; j < 9; j++){
                if(board[i][j] >= '0' && board[i][j] <= '9'){//橫著的
                    int a = board[i][j] - '0';
                    if(book[a] == 1)return false;
                    else            book[a] = 1;
                }
            }
            memset(book,0,sizeof(book));
            for(int j = 0; j < 9; j++){
                if(board[j][i] >= '0' && board[j][i] <= '9'){//豎著的
                    int a = board[j][i] - '0';
                    if(book[a] == 1)return false;
                    else            book[a] = 1;
                }
            }
            memset(book,0,sizeof(book));
        }
        for(int i = 0; i < 9; i += 3){
            for(int j = 0; j < 9; j += 3){
                memset(book,0,sizeof(book));
                for(int k = 0; k < 3; k++){//每個 3 * 3的矩形
                    for(int l = 0; l < 3; l++){
                        if(board[i + k][j + l] >= '0' && board[i + k][j + l] <= '9'){
                        int a = board[i + k][j + l] - '0';
                        if(book[a] == 1)return false;
                        else            book[a] = 1;
                        }
                    }
                }
            }
        }
    return true;
    }
    
};