【LeetCode】37. Sudoku Solver

Sudoku Solver

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.

这题跟N-Queens是一个套路，回溯法尝试所有解。

需要注意的区别是：

本题找到解的处理是return true，因此返回值为bool

N-Queen找到解的处理是保存解，因此返回值为void

对于每个空位'.'，遍历1~9，check合理之后往下一个位置递归。

由于这里路径尝试本质上是有序的，即1~9逐个尝试，因此无需额外设置状态位记录已经尝试过的方向。

注意：只有正确达到最终81位置（即成功填充）的填充结果才可以返回，若不然，将会得到错误的填充。

因此辅助函数solve需要设为bool而不是void

class Solution {
public:
    void solveSudoku(vector<vector<char> > &board) {
        solve(board, );
    }
    bool solve(vector<vector<char> > &board, int position)
    {
        if(position == )
            return true;

        int row = position / ;
        int col = position % ;
        if(board[row][col] == '.')
        {
            for(int i = ; i <= ; i ++)
            {//try each digit
                board[row][col] = i + '';
                if(check(board, position))
                    if(solve(board, position + ))
                    //only return valid filling
                        return true;
                board[row][col] = '.';
            }
        }
        else
        {
            if(solve(board, position + ))
            //only return valid filling
                return true;
        }
        return false;
    }
    bool check(vector<vector<char> > &board, int position)
    {
        int row = position / ;
        int col = position % ;
        int gid;
        if(row >=  && row <= )
        {
            if(col >=  && col <= )
                gid = ;
            else if(col >=  && col <= )
                gid = ;
            else
                gid = ;
        }
        else if(row >=  && row <= )
        {
            if(col >=  && col <= )
                gid = ;
            else if(col >=  && col <= )
                gid = ;
            else
                gid = ;
        }
        else
        {
            if(col >=  && col <= )
                gid = ;
            else if(col >=  && col <= )
                gid = ;
            else
                gid = ;
        }

        //check row, col, subgrid
        for(int i = ; i < ; i ++)
        {
            //check row
            if(i != col && board[row][i] == board[row][col])
                return false;

            //check col
            if(i != row && board[i][col] == board[row][col])
                return false;

            //check subgrid
            int r = gid/*+i/;
            int c = gid%*+i%;
            if((r != row || c != col) && board[r][c] == board[row][col])
                return false;
        }
        return true;
    }
};

check的另一种实现方式如下：

bool check(vector<vector<char> > &board, int pos)
    {
        int v = pos/;
        int h = pos%;
        char target = board[v][h];
        //row
        for(vector<char>::size_type st = ; st < ; st ++)
        {
            if(st != h)
            {
                if(target == board[v][st])
                    return false;
            }
        }

        //col
        for(vector<char>::size_type st = ; st < ; st ++)
        {
            if(st != v)
            {
                if(target == board[st][h])
                    return false;
            }
        }

        //subgrid
        int beginx = v/*;
        int beginy = h/*;
        for(int i = beginx; i < beginx+; i ++)
        {
            for(int j = beginy; j < beginy+; j ++)
            {
                if(i != v && j != h)
                {
                    if(target == board[i][j])
                        return false;
                }
            }
        }

        return true;
    }