【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.

class Solution {
public:
    bool isValidSudoku(const vector<vector<char>>& board) {
        bool used[9];
        for (int i = 0; i < 9; ++i) {
            memset(used, false, 9);
            for (int j = 0; j < 9; ++j) {
                if (board[i][j] == '.') continue;
                if (used[board[i][j] - '1']) {
                    return false;
                } else {
                    used[board[i][j] - '1'] = true;
                }
            }
            memset(used, false, 9);
            for (int j = 0; j < 9; ++j) {
                if (board[j][i] == '.') continue;
                if (used[board[j][i] - '1']) {
                    return false;
                } else {
                    used[board[j][i] - '1'] = true;
                }
            }
        }
        for (int r = 0; r < 3; ++r) {
            for (int c = 0; c < 3; ++c) {
                memset(used, false, 9);
                for (int i = 0; i < 3; ++i) {
                    for (int j = 0; j < 3; ++j) {
                        int v1 = r * 3 + i;
                        int v2 = c * 3 + j;
                        if (board[v1][v2] == '.') continue;
                        if (used[board[v1][v2] - '1']) {
                            return false;
                        } else {
                            used[board[v1][v2] - '1'] = true;
                        }
                    }
                }
            }
        }
        return true;
    }
};

依次检查即可，开始自作聪明想要在一个大的循环里面把三种方式都检查了，但是发现在检查小方块时下标运算会有除法和求模，反而使得程序变慢，另列一个3*3的循环虽然使代码变得复杂一点，但是思路清晰，且速度更快。