[leetcode] 20. Valid Sudoku

这道题目被放在的简单的类别里是有原因的，题目如下：

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.

又是一开始犯了审题错误的毛病，我以为是要写出一个计算数独的AI，还在想为啥这种难度都被认为成简单，结果我们看最后的note：

A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

就是说不需要这个数独是可以解，只要判断给你的这个数独题目是不是合法的数独就行了。所以难度就骤降了，随便找了一个解法上去AC。

class Solution {
public:
    bool isValidSudoku(vector<vector<char> > &board) {
        vector<vector<int>> bucket(3, vector<int>(9, 0));
         for (int i = 0; i < 9; i++)
         {
             for (int j = 0; j < 9; j++)
             {
                 if (board[i][j] != '.' && ++bucket[0][board[i][j] - '1'] > 1) return false;
                 if (board[j][i] != '.' && ++bucket[1][board[j][i] - '1'] > 1) return false;
                 int v = j%3 + 3*(i%3);
                 int h = j/3 + 3*(i/3);
                 if (board[v][h] != '.' && ++bucket[2][board[v][h] - '1'] > 1) return false;
             }
             bucket = vector<vector<int>>(3, vector<int>(9, 0));
         }
    }
};

每个点的合法性判断有三种，一种是横向是否有相同数字，一种是纵向是否有相同数字，一种是这个点所在的3*3格子内是否用相同数字。