【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[9];
        for (int i = 0; i < 9; ++i) {
            memset(count, 0, sizeof(int) * 9);
            for (int j = 0; j < 9; ++j) {
                if (!add(count, board[i][j]))
                    return false;
            }
        }
        return true;
    }
    
    bool isValidColumn(vector<vector<char> > board) {
        int count[9];
        for (int i = 0; i < 9; ++i) {
            memset(count, 0, sizeof(int) * 9);
            for (int j = 0; j < 9; ++j) {
                if (!add(count, board[j][i]))
                    return false;
            }
        }
        return true;
    }
    
    bool isValidBox(vector<vector<char> > board) {
        int count[9];
        int point[9][2] = {
            {0, 0}, {0, 3}, {0, 6}, {3, 0}, {3, 3}, {3, 6}, {6, 0}, {6, 3}, {6, 6}
        };
        for (int i = 0; i < 9; ++i) {
            memset(count, 0, sizeof(int) * 9);
            for (int x = 0; x < 3; ++x) {
                for (int y = 0; y < 3; ++y) {
                    int abscissa = point[i][0] + x;
                    int ordinate = point[i][1] + 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 - '1']) == 1;
    }
    
};

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