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.

题目要求：判断当前非'.'元素是否满足数独的行、列和方框要求。

解题思路：遍历数独，判断所有行、列是否都满足没有重复数字的要求，然后判断每个方框是否满足没有重复数字的要求。代码如下：

class Solution {
public:
    bool isValidSudoku(vector<vector<char> > &board) {  
        bool row[9],col[9];
        for(int i=0;i<9;i++)//判断每一行和列是否满足没有重复数字的要求
        {
                 memset(row,false,sizeof(bool)*9);
                 memset(col,false,sizeof(bool)*9);
            for(int j=0;j<9;j++)
            {
                 
                if(board[i][j]!='.')
                {
                    if(row[board[i][j]-'1'])return false;
                else row[board[i][j]-'1']=true;
                }
                 if(board[j][i]!='.')
                 {
                             if(col[board[j][i]-'1'])return false;
                else col[board[j][i]-'1']=true;
                 }
                
        
            }
            
        }
        for(int i=0;i<9;i+=3)//判断每个方框（共9个）是否满足没有重复数字的要求
        {
            for(int j=0;j<9;j+=3)
            {
                memset(row,false,sizeof(bool)*9);
                for(int a=0;a<3;a++)
                {
                    for(int b=0;b<3;b++)
                    {
                        if(board[i+a][j+b]=='.')continue;
                        if(row[board[i+a][j+b]-'1'])return false;
                        else row[board[i+a][j+b]-'1']=true;
                    }
                }
            }
        }
        return true;
    }  
};