題目：
判斷一個 9x9 的數(shù)獨是否有效。只需要根據(jù)以下規(guī)則，驗證已經(jīng)填入的數(shù)字是否有效即可。

數(shù)字 1-9 在每一行只能出現(xiàn)一次。
數(shù)字 1-9 在每一列只能出現(xiàn)一次。
數(shù)字 1-9 在每一個以粗實線分隔的 3x3 宮內(nèi)只能出現(xiàn)一次。

上圖是一個部分填充的有效的數(shù)獨。

數(shù)獨部分空格內(nèi)已填入了數(shù)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(shù)字從 5 改為 8 以外，空格內(nèi)其他數(shù)字均與 示例1 相同。
但由于位于左上角的 3x3 宮內(nèi)有兩個 8 存在, 因此這個數(shù)獨是無效的。
說明:

一個有效的數(shù)獨（部分已被填充）不一定是可解的。
只需要根據(jù)以上規(guī)則，驗證已經(jīng)填入的數(shù)字是否有效即可。
給定數(shù)獨序列只包含數(shù)字 1-9 和字符 '.' 。
給定數(shù)獨永遠是 9x9 形式的。

題目來源：https://leetcode-cn.com/problems/valid-sudoku/

思路：
1、按行統(tǒng)計每行是否有重復的數(shù)字
2、按列統(tǒng)計每列是否有重復的數(shù)字
3、按小九宮格統(tǒng)計是否有重復的數(shù)字：這里有個小技巧，通過判斷列所在的索引和3、6之間的關系來遞增的獲取小九宮格左上角位置的索引位置

C++代碼

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
    
        int flag[10];

        // 按行進行排查，看每行是否有重復的數(shù)字
        for(int i=0; i<9; i++){
            memset(flag, 0, sizeof(int)*9);  // 重置 flag
            for(int j=0; j<9; j++){
                if (board[i][j]!='.'){
                    if (flag[board[i][j]-'1']){
                        return false;
                    }else{
                        flag[board[i][j]-'1'] = 1;
                    }
                }
            }
        }

        // 按列進行排查，看每列是否有重復的數(shù)字
        for(int i=0; i<9; i++){
            memset(flag, 0, sizeof(int)*9);   // 重置 flag
            for(int j=0; j<9; j++){
                if (board[j][i]!='.'){
                    if (flag[board[j][i]-'1']){
                        return false;
                    }else{
                        flag[board[j][i]-'1'] = 1;
                    }
                }
            }
        }   

        for(int col=0, row=0; row<9; ){
            memset(flag, 0, sizeof(int)*9);   // 重置 flag
            for(int i=0; i<3; i++){
                for(int j=0; j<3; j++){
                    if (board[col+i][row+j]!='.'){
                        if (flag[board[col+i][row+j]-'1']){
                            return false;
                        }else{
                            flag[board[col+i][row+j]-'1'] = 1;
                        }
                    }
                }
            }  

            if (col == 6){
                col = 0;
                row += 3;
            }else{
                col += 3;
            }
        }

        return true;

    }
};

Python版：

    class Solution(object):
        def isValidSudoku(self, board):
            """
            :type board: List[List[str]]
            :rtype: bool
            """
            ret = True

            for i in range(9):
                dt = {}
                for j in range(9):
                    if board[i][j] != '.':
                        if board[i][j] in dt:
                            return False
                        else:
                            dt[board[i][j]] = 1

            for i in range(9):
                dt = {}
                for j in range(9):
                    if board[j][i] != '.':
                        if board[j][i] in dt:
                            return False
                        else:
                            dt[board[j][i]] = 1
            col = 0
            row = 0
            while row<9:
                dt = {}
                for i in range(3):
                    for j in range(3):
                        if board[col+i][row+j] != '.':
                            if board[col+i][row+j] in dt:
                                return False
                            else:
                                dt[board[col+i][row+j]] = 1

                if col == 6:
                    col = 0
                    row += 3
                else:
                    col += 3

            return ret