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。

 1 /*
 2     Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules:http://sudoku.com.au/TheRules.aspx.
 3     The Sudoku board could be partially filled, where empty cells are filled with the character '.'.
 4     Note:
 5     A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.
 6 
 7     Copyright ©2014 Vincent Zhang 
 8     Blog: http://www.centvin.com 
 9 */
10 class Solution {
11 public:
12     bool isValidSudoku(vector<vector<char> > &board) {
13         for (int row=0; row< BOARD_SIZE; row++) {
14             for (int col=0; col<BOARD_SIZE; col++) {
15                 if (board[row][col] != '.') {
16                     if (!isValid(board, row, col))
17                         return false;
18                 }
19             }
20         }
21         return true;
22     }
23 private:
24     bool isValid(vector<vector<char> > &board, int row, int col) {
25         char curChar = board[row][col];
26         // check the row and col
27         for (int i=0;i<BOARD_SIZE;i++) {
28             if (curChar == board[row][i] && i != col)        // row check
29                 return false;
30             if (curChar == board[i][col] && i != row)        // col check
31                 return false;
32         }
33         // check the sub block
34         int blockStartRow = (row / 3)*3;
35         int blockStartCol = (col / 3)*3;
36 
37         for (int i=blockStartRow;i<blockStartRow+3;i++) {
38             for (int j=blockStartCol; j<blockStartCol+3; j++) {
39                 if (curChar == board[i][j] && i != row && j != col) 
40                     return false;
41             }
42         }
43         return true;
44     }
45     static const int BOARD_SIZE = 9;
46     static const int BLOCK_SIZE = 3;
47 };