C++算法：有效的数独

leetcode相关C++算法解答: https://github.com/Nereus-Minos/C_plus_plus-leetcode

题目：

判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。
数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

示例 1:

	输入:
	[
	  ["5","3",".",".","7",".",".",".","."],
	  ["6",".",".","1","9","5",".",".","."],
	  [".","9","8",".",".",".",".","6","."],
	  ["8",".",".",".","6",".",".",".","3"],
	  ["4",".",".","8",".","3",".",".","1"],
	  ["7",".",".",".","2",".",".",".","6"],
	  [".","6",".",".",".",".","2","8","."],
	  [".",".",".","4","1","9",".",".","5"],
	  [".",".",".",".","8",".",".","7","9"]
	]
	输出: true

示例 2:

	输入:
	[
	  ["8","3",".",".","7",".",".",".","."],
	  ["6",".",".","1","9","5",".",".","."],
	  [".","9","8",".",".",".",".","6","."],
	  ["8",".",".",".","6",".",".",".","3"],
	  ["4",".",".","8",".","3",".",".","1"],
	  ["7",".",".",".","2",".",".",".","6"],
	  [".","6",".",".",".",".","2","8","."],
	  [".",".",".","4","1","9",".",".","5"],
	  [".",".",".",".","8",".",".","7","9"]
	]
	输出: false
	解释: 除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

方法一（三次迭代）：

分别写三个验证函数，行验证，列验证，3*3验证

方法二（一次迭代）：

使用三个存放map的vecotr来一次遍历

代码：

#if 0
//方法一：分别写三个验证函数，行验证，列验证，3*3验证
#if 0
//方法一：分别写三个验证函数，行验证，列验证，3*3验证
class Solution {
public:
    bool isValidSudoku(vector>& board) {
        //写三个函数分别验证
        return rowIsValid(board) && colIsValid(board) && threeIsValid(board);
    }
    
private:
    bool rowIsValid(vector>& board){
        
        for(int i = 0; i < 9; i++)
        {
            map temp;
            for(int j = 0; j < 9; j++)
            {
                if(board[i][j] != '.')
                {
                    if(temp.find(board[i][j]) != temp.end())
                        return false;
                    else
                        temp[board[i][j]] = 1;
                }
            }
        }
        return true;
    }
    
    bool colIsValid(vector>& board){
        for(int i = 0; i < 9; i++)
        {
            map temp;
            for(int j = 0; j < 9; j++)
            {
                if(board[j][i] != '.')
                {
                    if(temp.find(board[j][i]) != temp.end())
                        return false;
                    else
                        temp[board[j][i]] = 1;
                }
            }
        }
        return true;
    }
    
    bool threeIsValid(vector>& board){
        
        for(int i = 0; i < 3; i++)
        {
            for(int j = 0; j < 3; j++)
            {
                map temp;
                
                for(int k = 0; k < 9; k++)
                {
                    if(board[i*3+(k/3)][j*3+(k%3)] != '.')
                    {
                        if(temp.find(board[i*3+(k/3)][j*3+(k%3)]) != temp.end())
                            return false;
                        else
                            temp[board[i*3+(k/3)][j*3+(k%3)]] = 1;
                    }
                }
            }
        }
        return true;
    }
};
#endif

#if 1
//方法二：使用三个存放map的vecotr来一次遍历
class Solution {
public:
    bool isValidSudoku(vector>& board) {
        
        vector> row(9), col(9), sub(9);
        
        for(int i = 0; i < 9; i++)
        {
            for(int j = 0; j < 9; j++)
            {
                if(board[i][j] == '.')
                    continue;
                
                if(row[i].find(board[i][j]) != row[i].end())  //表明行中元素有重复
                    return false;
                   
                if(col[j].find(board[i][j]) != col[j].end())   //表明列中元素有重复
                    return false;
                
                if(sub[(i/3)*3 + j/3].find(board[i][j]) != sub[(i/3)*3 + j/3].end())  //表明3*3有重复
                    return false;
                
                row[i][board[i][j]] = 1;
                col[j][board[i][j]] = 1;
                sub[(i/3)*3 + j/3][board[i][j]] = 1;
            }
        }
       return true;
    }
    
};
#endif