LeetCode Valid Sudoku

思路：

对某一个有数字的格子检查是否valid，就是从行、列、宫三个方面来检查，检查的标准就是除了该格子，行、列、宫内的其他8个格子是否包含这个数字。那么实现就用两个循环，来逐行逐列的遍历，如果对应的格子里面有数字，那么就进行检查，否则扫描下一个格子。

/*
    Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules:http://sudoku.com.au/TheRules.aspx.
    The Sudoku board could be partially filled, where empty cells are filled with the character '.'.
    Note:
    A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

    Copyright ©2014 Vincent Zhang 
    Blog: http://www.centvin.com 
*/
class Solution {
public:
    bool isValidSudoku(vector<vector<char> > &board) {
        for (int row=0; row< BOARD_SIZE; row++) {
            for (int col=0; col<BOARD_SIZE; col++) {
                if (board[row][col] != '.') {
                    if (!isValid(board, row, col))
                        return false;
                }
            }
        }
        return true;
    }
private:
    bool isValid(vector<vector<char> > &board, int row, int col) {
        char curChar = board[row][col];
        // check the row and col
        for (int i=0;i<BOARD_SIZE;i++) {
            if (curChar == board[row][i] && i != col)        // row check
                return false;
            if (curChar == board[i][col] && i != row)        // col check
                return false;
        }
        // check the sub block
        int blockStartRow = (row / 3)*3;
        int blockStartCol = (col / 3)*3;

        for (int i=blockStartRow;i<blockStartRow+3;i++) {
            for (int j=blockStartCol; j<blockStartCol+3; j++) {
                if (curChar == board[i][j] && i != row && j != col) 
                    return false;
            }
        }
        return true;
    }
    static const int BOARD_SIZE = 9;
    static const int BLOCK_SIZE = 3;
};