LeetCode | 36. 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.

思路：判断每一行，每一列，每一个块里面是否有重复数字即可。行与列一起判断可以更快一些，16ms->15ms。
注：判重用 set 集合比较方便。

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        set<char> s;    //使用 set 方便查重
        set<char> s2;

        //对每一行和每一列检查
        for(int i=0;i<9;i++)
        {
            s.clear();
            s2.clear();
            for(int j=0;j<9;j++)
            {
                //每一行
                if(board[i][j]>='0' && board[i][j]<='9')
                {
                    if(s.find(board[i][j]) != s.end())
                        return false;       //有重合
                    s.insert(board[i][j]);
                }
                //每一列
                if(board[j][i]>='0' && board[j][i]<='9')
                {
                    if(s2.find(board[j][i]) != s2.end())
                        return false;       //有重合
                    s2.insert(board[j][i]);
                }
            }
        }

        //对每个块进行检查
        for(int i=0;i<7;i+=3)
        {
            for(int j=0;j<7;j+=3)
            {
                s.clear();
                for(int row=i;row<i+3;row++)
                {
                    for(int col=j;col<j+3;col++)
                    {
                        if(board[row][col]>='0' && board[row][col]<='9')
                        {
                            if(s.find(board[row][col]) != s.end())
                                return false;       //有重合
                            s.insert(board[row][col]);
                        }                        
                    }
                }
            }
        }

        return true;
    }
};