LeetCode（36）Valid Sudoku

题目

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.

分析

这是一道关于数独游戏的题目，首先要了解数独游戏的规则：

所以，对于该题目，有些空格中是’.’ 字符，我们只需要考虑当前状态下是否满足数独即可。
也就是说，我们要按行、按列，按每个3*3宫格，检验三次。

AC代码

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        if (board.empty())
            return false;

        //数独游戏符合9宫格，也就是为一个9*9的矩阵
        int size = board.size();

        //根据数独游戏的规则，需要进行三次检验（按行、按列、按照3*3块）
        //利用哈希的思想，记录每个关键字的出现次数，若是次数>1，则返回false
        vector<int> count;
        for (int i = 0; i < size; ++i)
        {
            //每行开始前，将记录次数的vector清零,元素1~9分别对应下标0~8,对应vector中值为该元素的出现次数
            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];
                }
                else
                    continue;

            }//for
        }//for

        //同理，按列检验
        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];;
                }
                else
                    continue;
            }//for
        }//for

        //按3*3小块检验
        for (int i = 0; i < size; i += 3)
        {           
            for (int j = 0; j < size; j += 3)
            {
                count.assign(9, 0);
                //每个块又是一个3*3的矩阵
                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;
                }
            }//for
        }//for

        return true;
    } 
};

GitHub测试程序源码