36. Valid Sudoku

problem:

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 '.'.

该问题是判断该数独盘是否合理，不用考虑最后能不能求解。

考虑3个方便。第一是每行没有数字重复，第二是每列没有数字重复，第三是每个3*3的方框里面无数字重复。

class Solution {public:    bool isValidSudoku(vector<vector<char>>& board) {        if (board.empty())            return false;                int size = board.size();        vector<int> count;        //考虑每行        for (int i = 0; i < size; ++i)        {            count.assign(9, 0);            for (int j = 0; j < size; j++)            {                if (board[i][j] != '.')                {                    int pos = board[i][j] - '1';                    if (count[pos] > 0)                        return false;                    else                        ++count[pos];                }                continue;            }        }        //考虑每列        for (int j = 0; j < size; ++j)        {            count.assign(9, 0);            for (int i = 0; i < size; i++)            {                if (board[i][j] != '.')                {                    int pos = board[i][j] - '1';                    if (count[pos] > 0)                        return false;                    else                        ++count[pos];                }                continue;            }        }        //考虑每3*3的方块内        for (int i = 0; i < size; i += 3)        {                       for (int j = 0; j < size; j += 3)            {                count.assign(9, 0);                for (int row = i; row < i + 3;row++)                for (int col = j; col < j + 3; col++)                {                    if (board[row][col] != '.')                    {                        int pos = board[row][col] - '1';                        if (count[pos] > 0)                            return false;                        else                            ++count[pos];;                    }                    else                        continue;                }            }        }        return true;    }};