LeetCode 37. Sudoku Solver

LeetCode 37. Sudoku Solver

问题

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.

分析

根据题目，我们可以获得以下信息：

• 这是个数独问题。即同一行同一列和同一个三阶方阵内的数字不能重复。

• 假设答案只有一个。即我们要用到深度优先搜索，找到答案后及时回溯。

根据这两个特点我采用了以下策略：

• 用递归的方法从左到右从上到下，逐渐完善这张图

• 如果得到最终答案，则停止递归。

• 如果这张图的下一个空位得不到任何一个符合要求的数字，则停止这次递归。

• 用一个函数实现递归。

• 用一个函数实现得到符合要求的数字列表。

实现代码

class Solution {public:    void solveSudoku(vector<vector<char>>& board) {        vector<vector<char>> result;        solveSudoku_iterator(0,0,board, result);        board = result;    }private:    void solveSudoku_iterator(int start_i,int start_j,vector<vector<char>> board, vector<vector<char>>& result) {        if (!result.empty())return;        int j = start_j;        for (int i = start_i; i < 9; i++)        {            for (; j < 9; j++)            {                if (board[i][j] == '.')                {                    vector<int> num;                    get_num(board, i, j,num);                    if (num.empty())return;                    while (!num.empty())                    {                        board[i][j] = 48 + num.back();                        solveSudoku_iterator(i,j, board, result);                        num.pop_back();                    }                    return;                }            }            j = 0;        }        result = board;    }    void get_num(vector<vector<char>>& board,int& x,int& y, vector<int>& num)    {        for (int n = 1 ; n < 10; n++)        {            bool pass = true;            for (int r_index = 0; r_index < 9 && pass; r_index++)            {                if (board[r_index][y] == n + 48 && r_index != x)pass = false;            }            for (int c_index = 0; c_index < 9 && pass; c_index++)            {                if (board[x][c_index] == n + 48 && c_index != y)pass = false;            }            for (int x_mod = (x / 3) * 3; x_mod < (x / 3 + 1)* 3 && pass; x_mod++)                for (int y_mod = (y / 3) * 3; y_mod < (y / 3 + 1) * 3 && pass; y_mod++)                    if (board[x_mod][y_mod] == n + 48 && (x_mod != x || y_mod != y))pass = false;            if (pass)num.push_back(n);        }    }};