[LettCode]49. Valid Sudoku有效数独

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.

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解法1：分别判断每行，每列，每个3X3块中是否存在相同数字，时间复杂度O(n^2)，空间复杂度O(n)。

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        if(board.empty() || board[0].empty()) return false;
        int n = board.size();
        vector<int> row(9, 0);
        vector<int> col(9, 0);
        for(int i = 0; i < n; ++i)
        {
            row.assign(9, 0);
            col.assign(9, 0);
            for(int j = 0; j < n; ++j)
            {
                if(board[i][j] != '.')
                {
                    if(row[board[i][j] - '1'] == 1) return false;
                    else ++row[board[i][j] - '1'];
                }
                if(board[j][i] != '.')
                {
                    if(col[board[j][i] - '1'] == 1) return false;
                    else ++col[board[j][i] - '1'];
                }
            }
        }
        vector<int> c3(9, 0);
        for(int k = 0; k < n * n / 9; ++k)
        {
            c3.assign(9, 0);
            int starti = k / 3 * 3, startj = k % 3 * 3;
            for(int col = starti; col < starti + 3; ++col)
            {
                for(int row = startj; row < startj + 3; ++row)
                {
                    if(board[col][row] != '.')
                    {
                        if(c3[board[col][row] - '1'] == 1) return false;
                        else ++c3[board[col][row] - '1'];
                    }
                }
            }
        }
        return true;
    }
};

在空间复杂度为O(n^2)时可以用一次二层循环解决：

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        if(board.empty() || board[0].empty()) return false;
        int n = board.size();
        vector<int> row(9, 0);
        vector<int> col(9, 0);
        vector< vector<int> > cub(9, vector<int>(9, 0));
        for(int i = 0; i < n; ++i)
        {
            row.assign(9, 0);
            col.assign(9, 0);
            for(int j = 0; j < n; ++j)
            {
                if(board[i][j] != '.')
                {
                    if(row[board[i][j] - '1'] == 1) return false;
                    else ++row[board[i][j] - '1'];
                    
                    int k = i / 3 * 3 + j / 3;
                    if(cub[k][board[i][j] - '1'] == 1) return false;
                    else ++cub[k][board[i][j] - '1'];
                }
                if(board[j][i] != '.')
                {
                    if(col[board[j][i] - '1'] == 1) return false;
                    else ++col[board[j][i] - '1'];
                }
            }
        }
        return true;
    }
};