LeetCode36——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.

难度系数:

容易

实现

bool isValidSudoku(vector<vector<char> > &board) {
    int bsudo[9];
    for (int i = 0; i < board.size(); ++i) {
        vector<char> &vc = board[i];
        memset(bsudo, 0, sizeof(int)*9);
        for (int j = 0; j < vc.size(); ++j) {
            if (vc[j] == '.')
                continue;
            if (bsudo[vc[j] - '1'] != 0)
                return false;
            bsudo[vc[j] - '1'] = 1;
        }
    }
    for (int i = 0; i < board[0].size(); ++i) {
        memset(bsudo, 0, sizeof(int)*9);
        for (int j = 0; j < board.size(); ++j) {
            if (board[j][i] == '.')
                continue;
            if (bsudo[board[j][i] - '1'] != 0)
                return false;
            bsudo[board[j][i] - '1'] = 1;
        }
    }
    int m = 0;

    for (int i = 0; i < 3; i++,m += 3) {
        int n = 0;
        for (int j = 0; j < 3; j++,n += 3) {
            memset(bsudo, 0, sizeof(int)*9);
            for (int p = 0; p < 3; p++) {
                for (int q = 0; q < 3; q++) {
                    if (board[m+p][n+q] == '.')
                        continue;
                    if (bsudo[board[m+p][n+q] - '1'] != 0) {
                        return false;
                    }
                    bsudo[board[m+p][n+q] - '1'] = 1;
                }
            }
        }
    }
    return true;
}

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