【leetcode】Valid Sudoku

Valid Sudoku

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

A partially filled sudoku which is valid.

Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

数独的规则：

在一个9*9的区域内，

每行1-9出现且只出现一次，

每列1-9出现且只出现一次，

在9个子3*3的区域内1-9出现且只出现一次。

 这题允许填空('.')，所以不必要每个数字都出现。

class Solution {
public:
    bool isValidSudoku(vector<vector<char> > &board) {
        
        return  isValidRow(board)&&isValidColumn(board)&&isValidBox(board);
    }
    
    
    bool isValidRow(vector<vector<char> > &board)
    {
        map<int,bool> hash;
        for(int i=0;i<9;i++)
        {
            for(int j=0;j<9;j++)
            {
                if(board[i][j]=='.') continue;
                if(hash.find(board[i][j])!=hash.end()) return false;
                hash[board[i][j]]=true;
                
            }
            hash.clear();
        }
        return true;
    }
    
    bool isValidColumn(vector<vector<char> > &board)
    {
        map<int,bool> hash;
        for(int i=0;i<9;i++)
        {
            for(int j=0;j<9;j++)
            {
                if(board[j][i]=='.') continue;
                if(hash.find(board[j][i])!=hash.end()) return false;
                hash[board[j][i]]=true;
                
            }
            hash.clear();
        }
        return true;
    }
    
    bool isValidBox(vector<vector<char> > &board)
    {
        int p[9][2]={{0,0},{0,3},{0,6},{3,0},{3,3},{3,6},{6,0},{6,3},{6,6}};
        map<int,bool> hash;
        for(int h=0;h<9;h++)
        {
            for(int i=0+p[h][0];i<3+p[h][0];i++)
            {
                for(int j=0+p[h][1];j<3+p[h][1];j++)
                {
                    if(board[i][j]=='.') continue;
                    if(hash.find(board[i][j])!=hash.end()) return false;
                    hash[board[i][j]]=true;
                }
            }
            hash.clear();
        }
        return true;
    }
};