判断数独是否合法-LintCode

请判定一个数独是否有效。
该数独可能只填充了部分数字，其中缺少的数字用 . 表示。

注意事项：
一个合法的数独（仅部分填充）并不一定是可解的。我们仅需使填充的空格有效即可。

说明：
什么是 数独？
http://sudoku.com.au/TheRules.aspx
http://baike.baidu.com/subview/961/10842669.htm

样例：

The following partially filed sudoku is valid.

#ifndef C389_H#define C389_H#include<iostream>#include<vector>#include<map>using namespace std;class Solution {public:    /*    * @param board: the board    * @return: whether the Sudoku is valid    */    bool isValidSudoku(vector<vector<char>> &board) {        // write your code here        int m = board.size();        int n = board[0].size();        map<char, int> count;        for (int i = 0; i < m; ++i)        {            count.clear();            for (int j = 0; j < n; ++j)            {                if (isdigit(board[i][j]))                {                    if (count.find(board[i][j]) == count.end())                    {                        count[board[i][j]]++;                    }                    else                    {                        return false;                        break;                    }                }            }        }        for (int j = 0; j < n; ++j)        {            count.clear();            for (int i = 0; i < m; ++i)            {                if (isdigit(board[i][j]))                {                    if (count.find(board[i][j]) == count.end())                    {                        count[board[i][j]]++;                    }                    else                    {                        return false;                        break;                    }                }            }        }        vector<vector<int>> dir{ { 0, 0 }, { -1, 0 }, { -1, -1 }, { 0, -1 }, { 1, -1 }, { 1, 0 }, { 1, 1 }, { 0, 1 }, { -1, 1 } };        vector<vector<int>> center{ { 1, 1 }, { 1, 4 }, { 1, 7 }, { 4, 1 }, { 4, 4 }, { 4, 7 }, { 7, 1 }, { 7, 4 }, { 7, 7 } };        for (auto c : center)        {            count.clear();            for (auto t : dir)            {                int x = c[0] + t[0];                int y = c[1] + t[1];                if (isdigit(board[x][y]))                {                    if (count.find(board[x][y]) == count.end())                    {                        count[board[x][y]]++;                    }                    else                    {                        return false;                        break;                    }                }            }        }        return true;    }};#endif