131_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.

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

1:注意特殊情況；2:遍歷每一個非空的cell，檢測當前行或者當前列，或者所在的小正方形中是不是滿足數獨的條件，如果不滿足，直接返回false。遍歷所有的結果都滿足的話，返回true

    bool isValidSudoku(vector<vector<char> > &board)
    {
        if(board.size() != 9 || board[0].size() != 9)
        {
            return false;
        }
        
        int rows = (int)board.size();
        int columns = (int)board[0].size();
        
        for(int i = 0; i < rows; i++)
        {
            for(int j = 0; j < columns; j++)
            {
                if(board[i][j] == '.')
                {
                    continue;
                }
                else
                {
                    if(isValidSudokuCore(board, i, j))
                    {
                        continue;
                    }
                    else
                    {
                        return false;
                    }
                }
            }
        }
        return true;
    }
    
    bool isValidSudokuCore(vector<vector<char> > &board, int curX, int curY)
    {
        for(int i = 0; i < (int)board.size(); i++)
        {
            if(i != curX && board[i][curY] !='.' && board[i][curY] == board[curX][curY])
            {
                return false;
            }
        }
        
        for(int i = 0; i < (int)board[0].size(); i++)
        {
            if(i != curY && board[curX][i] != '.' && board[curX][i] == board[curX][curY])
            {
                return false;
            }
        }
        
        int rows = curX / 3;
        int columns = curY / 3;
        
        for(int i = 0; i < 3; i++)
        {
            for(int j = 0; j < 3; j++)
            {
                if((rows * 3 + i != curX || columns * 3 + j != curY) && board[rows * 3 + i][columns * 3 + j] != '.' && board[rows * 3 + i][columns * 3 + j] == board[curX][curY])
                {
                    return false;
                }
            }
        }
        
        return true;
    }