[Leetcode 103] 37 Sudoku Solver

Problem:

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.

Analysis:

The basic dfs problem. Notice the structure for DFS function. It's a well structured program abstraction.

Also notice when judging whether there's conflict in a sub 3*3 grid, the start point is i/3 * 3 and j/3*3. Please don't forget to multiple the second 3. Or that's not right

Code:

class Solution {
public:
    void solveSudoku(vector<vector<char> > &board) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        solve(board);
    }
    
private:
    bool solve(vector<vector<char> > &board) {
        int r, c;

        if (noRemainingPos(board, r, c))
            return true;

        for (int i=1; i<=9; i++) {
            if (noConflict(board, r, c, i+'0')) {
                board[r][c] = i+'0';   
                if (solve(board)) return true;
                board[r][c] = '.';
            }
        }
        
        return false;
    }
    
    bool noRemainingPos(const vector<vector<char> > &b, int &r, int &c) {
        for (int i=0; i<9; i++)
            for (int j=0; j<9; j++) {
                if (b[i][j] == '.') {
                    r = i;
                    c = j;
                    return false;
                }
                
            }
        
        return true;
    }
    
    bool noConflict(const vector<vector<char> > &b, int r, int c, char n) {
        return noConf(b, r, 0, 1, 9, n) && noConf(b, 0, c, 9, 1, n) && noConf(b, r/3*3, c/3*3, 3, 3, n);    
    }
    
    bool noConf(const vector<vector<char> > &b, int r, int c, int rc, int cc, char n) {
        for (int i=r; i<r+rc; i++)
            for (int j=c; j<c+cc; j++) {
                if (b[i][j] == n)
                    return false;
            }
        
        return true;
    }
    
};