36. 有效的数独

Posted 2022-12-02 　落禅

36. 有效的数独

请你判断一个 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。（请参考示例图）
数独部分空格内已填入了数字，空白格用 ‘.’ 表示。

注意：

一个有效的数独（部分已被填充）不一定是可解的。
只需要根据以上规则，验证已经填入的数字是否有效即可。

输入：board = 
[["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：board = 
[["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。

思路：这道题是判断这个数独是否有效，而题目已经告诉我们这是一个9x9的数独，那么我们只需要判断这个9x9的数独每一行，每一列这个数字是否只出来了一次，然后再判将9x9的数独分成9个3x3的小数独，判断这个3x3的格子里面每个数字是否只出现了一次，如果在每一行，每一列，每一个3x3的方格中每个数字都只出现了一次，那么就是有效的数独，反之就是无效的数独，那么这道题目的关键就变为：

1.该数字只在改行出现一次

2.该数字只在该列出现一次

3.该数字在各个3x3的放个里也只出现了一次

class Solution 
public:
//最粗暴的方法：循环遍历每一个字符，判断他在这一行是否只出现一次，在这里列只出现一次，在周围3*3的宫格内只出现了一次

    bool isValidSudoku(vector<vector<char>>& board) 
    //1.判断每个数字在改行只出现了一次
        for(int i=0;i<9;i++)
        
        	//定义一个一维数组，采用哈希的思想，将这一行映射到这个数组中去
            int row[9]=0;
            for(int j=0;j<9;j++)
            
            	//如果不是数字进入下一层循环
                if(board[i][j]=='.')
                continue;
                //如果(row[board[i][j]-'1']!=0，说明这个数已经在前面出现过一次了，直接返回false
                if(row[board[i][j]-'1']!=0)
                return false;
                
                row[board[i][j]-'1']=1;
            
        
        //2.判断每个数字在该列只出现了一次
        for(int i=0;i<9;i++)
        
              int col[9]=0;
            for(int j=0;j<9;j++)
            
                if(board[j][i]=='.')
                continue;
                if(col[board[j][i]-'1']!=0)
                return false;
                col[board[j][i]-'1']=1;
            
        
        //3.判断每个数字在该3x3的格子里也只出现了一次
        //将9x9的数独分割成9个3x3的小格子，然后像上面一样判断在这个格子里每个数字是否只出现了一次
        for(int i=0;i<3;i++)
        
            for(int j=0;j<3;j++)
            
                int check[9]=0;
                for(int ii=i*3;ii<i*3+3;ii++)
                
                    
                    for(int jj=j*3;jj<j*3+3;jj++)
                    
                        if(board[ii][jj]=='.')
                        continue;
                        if(check[board[ii][jj]-'1']!=0)
                        return false;
                        check[board[ii][jj]-'1']=1;
                    
                
            
        
        return true;
    
;

ps:由于上述循环都是常数级别的，因此时间复杂度为O(1)

感谢大家的观看，期待大家的评论，点赞，收藏，我们下次再见！ To Be Continued…