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);
}
}
};