【Leetcode】【Easy】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.

解题：

将每一行、每一列、每一个九格中的数字分别输入一个大小为9的数组中，记录输入的数字是否重复

代码：

 class Solution {
 public:
     bool isValidSudoku(vector<vector<char> > &board) {
         return isValidRow(board) && isValidColumn(board) && isValidBox(board);
     }

     bool isValidRow(vector<vector<char> > board) {
         int count[];
         for (int i = ; i < ; ++i) {
             memset(count, , sizeof(int) * );
             for (int j = ; j < ; ++j) {
                 if (!add(count, board[i][j]))
                     return false;
             }
         }
         return true;
     }

     bool isValidColumn(vector<vector<char> > board) {
         int count[];
         for (int i = ; i < ; ++i) {
             memset(count, , sizeof(int) * );
             for (int j = ; j < ; ++j) {
                 if (!add(count, board[j][i]))
                     return false;
             }
         }
         return true;
     }

     bool isValidBox(vector<vector<char> > board) {
         int count[];
         int point[][] = {
             {, }, {, }, {, }, {, }, {, }, {, }, {, }, {, }, {, }
         };
         for (int i = ; i < ; ++i) {
             memset(count, , sizeof(int) * );
             for (int x = ; x < ; ++x) {
                 for (int y = ; y < ; ++y) {
                     int abscissa = point[i][] + x;
                     int ordinate = point[i][] + y;
                     if (!add(count, board[abscissa][ordinate]))
                         return false;
                 }
             }
         }
         return true;
     }

     bool add(int count[], char dig) {
         if (dig == '.')
             return true;
         else
             return (++count[dig - '']) == ;
     }

 };

代码疑问：

1、为什么形参设置为board和&board都可以通过编译；

附录：

数独填充算法