Sudoku Solver

最新推荐文章于 2025-05-31 23:57:32 发布

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最新推荐文章于 2025-05-31 23:57:32 发布

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分类专栏： leetcode习题分析文章标签：数独递归回溯

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本文介绍了一种使用递归加回溯算法解决数独问题的方法。通过预先填充已知数字并利用递归逐个尝试未知格子的数字，确保每一行、每一列以及每一个小九宫格内的数字都不重复。

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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.

解法：递归加回溯

class Solution {
public:
    char row[9][9];  
    char column[9][9];  
    char block[9][9]; 
    
    void solveSudoku(vector<vector<char> > &board) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        makeMatrix(board);
        int flag = 0;
        solve(board,0,0,flag);
    }
    
    void solve(vector<vector<char>>& board,int i,int j,int& flag)
    {
        if(board[i][j] == '.'){
            for(char k = '1';k<='9';k++)
            {
                int digit = k-'1';  
                if(row[i][digit] != 0) continue;  
                if(column[j][digit] != 0) continue;  
                if(block[i/3 * 3 + j/3][digit] != 0) continue;
                row[i][digit] = 1;  
                column[j][digit] = 1;  
                block[i/3 * 3 + j/3][digit] = 1;  
         
           
                board[i][j] = k;
                
                if(j < 8) solve(board,i,j+1,flag);
                else if(i < 8) solve(board,i+1,0,flag);
                else flag = 1;
    			if(flag == 1) return;

                board[i][j] = '.';
                row[i][digit] = 0;  
                column[j][digit] = 0;  
                block[i/3 * 3 + j/3][digit] = 0;  
                
            }
        }else{
            if(j < 8) solve(board,i,j+1,flag);
            else if(i < 8) solve(board,i+1,0,flag);
            else flag = 1;
			if(flag == 1) return;
        }
    }
    
    void makeMatrix(vector<vector<char> > &board) {
        // Start typing your C/C++ solution berlow
        // DO NOT write int main() function
      
        memset(row,0,81);
        memset(column,0,81);
        memset(block,0,81);
        
        for(int i = 0;i<9;i++)
        {
            for(int j = 0;j<9;j++)
            {    
                if(board[i][j] == '.') continue;
                int digit = board[i][j]-'1';
                row[i][digit] = 1;
                column[j][digit] = 1;
                block[i/3 * 3 + j/3][digit] = 1;
            }
        }
    }
};

20 milli secs.