【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.

 

這題跟N-Queens是一個套路，回溯法嘗試所有解。

需要注意的區別是：

本題找到解的處理是return true，因此返回值為bool

N-Queen找到解的處理是保存解，因此返回值為void

 

對於每個空位'.'，遍歷1~9，check合理之后往下一個位置遞歸。

由於這里路徑嘗試本質上是有序的，即1~9逐個嘗試，因此無需額外設置狀態位記錄已經嘗試過的方向。

注意：只有正確達到最終81位置（即成功填充）的填充結果才可以返回，若不然，將會得到錯誤的填充。

因此輔助函數solve需要設為bool而不是void

class Solution {
public:
    void solveSudoku(vector<vector<char> > &board) {
        solve(board, 0);
    }
    bool solve(vector<vector<char> > &board, int position)
    {
        if(position == 81)
            return true;

        int row = position / 9;
        int col = position % 9;
        if(board[row][col] == '.')
        {
            for(int i = 1; i <= 9; i ++)
            {//try each digit
                board[row][col] = i + '0';
                if(check(board, position))
                    if(solve(board, position + 1))
                    //only return valid filling
                        return true;
                board[row][col] = '.';
            }
        }
        else
        {
            if(solve(board, position + 1))
            //only return valid filling
                return true;
        }
        return false;
    }
    bool check(vector<vector<char> > &board, int position)
    {
        int row = position / 9;
        int col = position % 9;
        int gid;
        if(row >= 0 && row <= 2)
        {
            if(col >= 0 && col <= 2)
                gid = 0;
            else if(col >= 3 && col <= 5)
                gid = 1;
            else
                gid = 2;
        }
        else if(row >= 3 && row <= 5)
        {
            if(col >= 0 && col <= 2)
                gid = 3;
            else if(col >= 3 && col <= 5)
                gid = 4;
            else
                gid = 5;
        }
        else
        {
            if(col >= 0 && col <= 2)
                gid = 6;
            else if(col >= 3 && col <= 5)
                gid = 7;
            else
                gid = 8;
        }

        //check row, col, subgrid
        for(int i = 0; i < 9; 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/3*3+i/3;
            int c = gid%3*3+i%3;
            if((r != row || c != col) && board[r][c] == board[row][col])
                return false;
        }
        return true;
    }
};

check的另一種實現方式如下：

bool check(vector<vector<char> > &board, int pos)
    {
        int v = pos/9;
        int h = pos%9;
        char target = board[v][h];
        //row
        for(vector<char>::size_type st = 0; st < 9; st ++)
        {
            if(st != h)
            {
                if(target == board[v][st])
                    return false;
            }
        }

        //col
        for(vector<char>::size_type st = 0; st < 9; st ++)
        {
            if(st != v)
            {
                if(target == board[st][h])
                    return false;
            }
        }

        //subgrid
        int beginx = v/3*3;
        int beginy = h/3*3;
        for(int i = beginx; i < beginx+3; i ++)
        {
            for(int j = beginy; j < beginy+3; j ++)
            {
                if(i != v && j != h)
                {
                    if(target == board[i][j])
                        return false;
                }
            }
        }

        return true;
    }