leetcode 36. Valid Sudoku

谁践踏了优雅 2022-07-26 05:14 112阅读 0赞

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 `'.'`.

![250px-Sudoku-by-L2G-20050714.svg.png][]

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.

class Solution {
    public:
    	bool isValidSudoku(vector<vector<char>>& board) {
    		char aa[9] = { '1', '2', '3', '4', '5', '6', '7', '8', '9' };
    		set <char>nums(aa, aa + 9);
    		for (int i = 0; i < board.size(); i++)
    		{
    			set<char>bb = nums;
    			for (int j = 0; j < board[0].size(); j++)
    				if (board[i][j] != '.')
    				{
    				set<char>::iterator it = bb.find(board[i][j]);
    				if (it == bb.end())
    					return false;
    				bb.erase(it);
    				}
    		}
    		for (int i = 0; i < board[0].size(); i++)
    		{
    			set<char>bb = nums;
    			for (int j = 0; j < board.size(); j++)
    				if (board[j][i] != '.')
    				{
    				set<char>::iterator it = bb.find(board[j][i]);
    				if (it == bb.end())
    					return false;
    				bb.erase(it);
    				}
    		}
    		for (int i = 0; i < board.size(); i += 3)
    		{
    			for (int j = 0; j < board[0].size(); j += 3)
    			{
    				set<char>bb = nums;
    				for (int y = 0; y < 3; y++)
    					for (int x = 0; x < 3; x++)
    						if (board[i + y][j + x] != '.')
    						{
    					set<char>::iterator it = bb.find(board[i + y][j + x]);
    					if (it == bb.end())
    						return false;
    					bb.erase(it);
    						}
    
    			}
    		}
    		return true;
    	}
    };

[Sudoku Puzzles - The Rules]: http://sudoku.com.au/TheRules.aspx
[250px-Sudoku-by-L2G-20050714.svg.png]: /images/20220724/ed5965d4ab0845a4885efd0dc9308098.png