Leetcode: 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.

就是每一行，每一列，每一个方块的看

bool isValidSudoku(vector<vector<char> > &board) {
        // Note: The Solution object is instantiated only once.
		int row[9],col[9];
		
		for(int i = 0; i < 9; i++)
		{
			memset(row,0,sizeof(int)*9);
			memset(col,0,sizeof(int)*9);
			for(int j = 0; j < 9; j++)
			{
				if(board[i][j] != '.')
				{
					if(row[board[i][j]-'1'] > 0)return false;
					else row[board[i][j]-'1']++;
				}
				if(board[j][i] != '.')
				{
					if(col[board[j][i]-'1'] > 0)return false;
					else col[board[j][i]-'1']++;
				}
			}
		}

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

简化版：

bool isValidSudoku(vector<vector<char> > &board) {
        // Note: The Solution object is instantiated only once.
		vector<vector<bool>> rows(9, vector<bool>(9,false));
		vector<vector<bool>> cols(9, vector<bool>(9,false));
		vector<vector<bool>> blocks(9, vector<bool>(9,false));

		for(int i = 0; i < 9; i++)
			for(int j = 0; j < 9; j++)
			{
				if(board[i][j] == '.')continue;
				int num = board[i][j] - '1';
				if(rows[i][num] || cols[j][num] || blocks[i - i%3 + j/3][num])
					return false;
				rows[i][num] = cols[j][num] = blocks[i - i%3 + j/3][num] = true;
			}
		return true;
    }