LeetCode之“散列表”：Valid Sudoku

　　题目链接

　　题目要求：

　　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.

　　这道题解法暂没有比较巧妙的，所以下边程序所用方法就是Brute Force（暴力破解）：

class Solution {
public:
    int charToInt(char c)
    {
        char str[] = {'1', '2', '3', '4', '5', '6', '7', '8', '9'};
        int intStr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
        for(int i = 0; i < 9; i++)
        {
            if(c == str[i])
                return intStr[i];
        }
        return -1;
    }
    
    bool isValidSudoku(vector<vector<char>>& board) {
        unordered_map<int, int> hashMap;
        for(int i = 1; i < 10; i++)
            hashMap[i] = 0;
        // row by row
        for(int i = 0; i < 9; i++)
        {
            for(int k = 0; k < 10; k++)
                hashMap[k] = 0;
            for(int j = 0; j < 9; j++)
            {
                int tmp = charToInt(board[i][j]);
                if(tmp != -1)
                {
                    hashMap[tmp]++;
                    if(hashMap[tmp] > 1)
                        return false;
                }
            }
        }
        // column by column
        for(int j = 0; j < 9; j++)
        {
            for(int k = 1; k < 10; k++)
                hashMap[k] = 0;
            for(int i = 0; i < 9; i++)
            {
                int tmp = charToInt(board[i][j]);
                if(tmp != -1)
                {
                    hashMap[tmp]++;
                    if(hashMap[tmp] > 1)
                        return false;
                }
            }
        }
        // 3*3 boxes by 3*3 boxes
        for(int i = 0; i < 9; i += 3)
        {
            for(int j = 0; j < 9; j += 3)
            {
                for(int k = 0; k < 10; k++)
                    hashMap[k] = 0;
                for(int m = i; m < i + 3; m++)
                    for(int n = j; n < j + 3; n++)
                    {
                        int tmp = charToInt(board[m][n]);
                        if(tmp != -1)
                        {
                            hashMap[tmp]++;
                            if(hashMap[tmp] > 1)
                                return false;
                        }
                    }
            }
        }
        
        return true;
    }
};