LeetCode: 37. Sudoku Solver

LeetCode: 37. Sudoku Solver

题目描述

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.

大意就是说: 已知某个有且仅有唯一解的数独, 让我们写个代码解开它。

解题思路

用 DFS 就可以做出了。这里我们利用 LeetCode: 36. Valid Sudoku 的思路, 对数独进行标记。

AC源码

class Solution {
public:
bool solveSudoku(vector<vector<char>>& board, int i, int j, int flag[][9])
{
if(i==9) return true;
if(j==9) return solveSudoku(board, i+1, 0, flag);

if(board[i][j] != '.')
{
return solveSudoku(board, i, j+1, flag);
}
else
{
for(int digit = 0; digit < 9; ++digit)
{
int f = flag[0][i] | flag[1][j] | flag[2][i/3*3+j/3];
if(f & (1 << digit))
{
continue;
}
else
{
// 标记
flag[0][i] |= (1 << digit);
flag[1][j] |= (1 << digit);
flag[2][i/3*3+j/3] |= (1 << digit);
board[i][j] = digit + '1';

if(solveSudoku(board, i, j+1, flag)) return true;

// 取消标记
board[i][j] = '.';
flag[0][i] &= ~(1 << digit);
flag[1][j] &= ~(1 << digit);
flag[2][i/3*3+j/3] &= ~(1 << digit);
}
}
}
return false;
}
void solveSudoku(vector<vector<char>>& board) {
int flag[3][9] = {0}; // 0: row, 1: column, 2: rect

// 初始化标记
for(int i = 0; i < board.size(); ++i)
{
for(int j = 0; j < board[i].size(); ++j){
if(board[i][j] != '.')
{
int value = 1 << (board[i][j] - '0' - 1);
if(flag[0][i] & value) return ;
else flag[0][i] |= value;

if(flag[1][j] & value) return ;
else flag[1][j] |= value;

int rectIdx = i/3*3 + j/3;
if(flag[2][rectIdx]&value) return ;
else flag[2][rectIdx] |= value;
}
}
}
solveSudoku(board, 0, 0, flag);
}
};