【LeetCode】Valid Sudoku-优快云博客

该博客讨论了如何验证一个部分填充的数独是否有效，根据数独谜题的规则，并提到了存在有效的数独板可能无法解决。此外，还提出了编写程序来解决数独谜题的问题，即填空白格，确保只有一个唯一的解决方案。

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参考链接

http://blog.youkuaiyun.com/doc_sgl/article/details/13002461

题目描述

Valid Sudoku

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.

这个题目并不是问有没有一种解决方案，而是问填进去的数字是否合理，只要保证填进去的数字一行，一列一大格里没重复数字就可以了。

bool isValidSudoku(vector<vector<char> > &board) {
        // Note: The Solution object is instantiated only once.
		vector<vector<bool>> rows(9, vector<bool>(9,false));
		vector<vector<bool>> cols(9, vector<bool>(9,false));
		vector<vector<bool>> blocks(9, vector<bool>(9,false));

		for(int i = 0; i < 9; i++)
			for(int j = 0; j < 9; j++)
			{
				if(board[i][j] == '.')continue;
				int num = board[i][j] - '1';
				if(rows[i][num] || cols[j][num] || blocks[i - i%3 + j/3][num])
					return false;
				rows[i][num] = cols[j][num] = blocks[i - i%3 + j/3][num] = true;
			}
		return true;
    }

Sudoku Solver

Total Accepted: 6798 Total Submissions: 33090 My Submissions

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle...

...and its solution numbers marked in red.

class Solution {
public:
    void solveSudoku(vector<vector<char> > &board) {
     	   solveSudokuCore(board,0);
    }
    bool solveSudokuCore(vector<vector<char> > &board,int index)
    {
    	if(index == 81) return true;
    	
    	int m = index / 9, n = index%9;
    	if(board[m][n] != '.')
    	{
    		return solveSudokuCore(board,index+1);
		}
    //	printvecvec(board,"board");	
    //	system("pause");
    	set<char> leftnum;
    	for(int i = 1;i<10;i++)
    		leftnum.insert((char)(i+'0'));
   		
   		for(int i = 0;i<9;i++)
   		{
   			if(board[m][i] != '.')
    			leftnum.erase(board[m][i]);
   		}
   		
   		for(int i = 0;i<9;i++)
   		{
   			if(board[i][n] != '.')
    			leftnum.erase(board[i][n]);
   		}
   		int r = (m/3)*3;
   		int c = (n/3)*3;
   		for(int i = 0;i<3;i++)
   			for(int j = 0;j<3;j++)
   			{
   				if(r+i != m && c+j != n && board[r+i][c+j] != '.')
   					leftnum.erase(board[r+i][c+j]);
		   	}
	   	set<char>::iterator it;
	   /*	printf("++%d,%d :",m,n);
	   	for (it=leftnum.begin(); it!=leftnum.end(); ++it)
   		{
				printf("%c  ", *it);
	   	}
	   	printf("\n");
   		*/
   		
   		for (it=leftnum.begin(); it!=leftnum.end(); ++it)
   		{
   			board[m][n] = *it;
   			//printf("%d,%d %c test\n",m,n,*it);
   			if(solveSudokuCore(board,index+1))
   				return true;
		//	else
		//		printf("%d,%d %c not ok\n",m,n,*it);
	   	}
	   	board[m][n] = '.';//////////////////////////////////////////////////////////////////
	   	return false;
    		
	}
};