LeetCode 37. Sudoku Solver

Posted 2022-12-08 hankunyan

经典backtracing的问题。我们可以记录每行，每列，每个box那些数字出现过，快速判断当前填入的数字是否有重复。

class Solution 
public:
    vector<vector<bool>> rowhash; // rowhash[row][1~9] - if the number has been taken in row
    vector<vector<bool>> colhash; // colhash[col][1~9] - if the number has been taken in col
    vector<vector<bool>> boxhash; // box are indexed by 0~8, from left to right, top to bottom
    
    void solveSudoku(vector<vector<char>>& board) 
        int m=board.size(), n=m==0?0:board[0].size();
        if (m==0 || n==0) return;
        
        rowhash.resize(9,vector<bool>(10,false));
        colhash.resize(9,vector<bool>(10,false));
        boxhash.resize(9,vector<bool>(10,false));
        
        for (int i=0;i<m;++i)
            for (int j=0;j<n;++j)
                if (board[i][j]==‘.‘) continue;
                int num=board[i][j]-‘0‘;
                rowhash[i][num] = true;
                colhash[j][num] = true;
                boxhash[i/3*3+j/3][num] = true; // i/3, j/3
            
        
        backtrack(board,0,0);
    
    
    bool backtrack(vector<vector<char>> &board, int i, int j)
        if (i==9) return true;
        if (board[i][j]!=‘.‘)
            auto [next_i,next_j] = nextPos(i,j);
            return backtrack(board,next_i,next_j);
        else
            // select a number to fill in the blank
            for (int k=1;k<=9;++k)
                if (!rowhash[i][k] && !colhash[j][k] && !boxhash[i/3*3+j/3][k])
                    board[i][j] = k+‘0‘;
                    rowhash[i][k] = colhash[j][k] = boxhash[i/3*3+j/3][k] = true;
                    auto [next_i,next_j] = nextPos(i,j);
                    if (backtrack(board,next_i,next_j))
                        return true;
                    board[i][j] = ‘.‘;
                    rowhash[i][k] = colhash[j][k] = boxhash[i/3*3+j/3][k] = false;
                
            
        
        return false;
    
    
    pair<int,int> nextPos(int i, int j)
        if (j==8) return i+1,0;
        else return i,j+1;
    
;

上述方法是dfs(i,j)，导致没到一行的末尾要换行，比较繁琐。

可以把所有空格都放到一个vector里，dfs这个vector的下标即可。