Leetcode: 37. Sudoku Solver

Description

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.

思路

非常常规的的一个dfs

借用36题中的bitset做映射，该位置0表未出现该数字，该位置1表出现

然后将9行，9列，9个小矩阵映射到27个bitset<9>上面

代码

class Solution {
public:
void solveSudoku(vector<vector<char>>& board) {
int m = board.size();
int n = board[0].size();
vector<bitset<9>> flag(27, 0);
int num = 0;
for (int i = 0; i < m; ++i){
for (int j = 0; j < n; ++j){
if (board[i][j] == '.') continue;
num = board[i][j] - '1';
flag[i].set(num);
flag[9 + j].set(num);
flag[18 + (i / 3) * 3 + j / 3].set(num);
}
}

solver(board, m, n, 0, 0, flag);
}

bool solver(vector<vector<char>> &board, int m, int n, int i, int j, vector<bitset<9>>& flag){

while (i < m){
if (j == n){
j = 0;
i++;
}

if (i < m && j < n && board[i][j] != '.'){
j++;
continue;
}
else break;
}

if (i == m) return true;

for (int t = 0; t < 9; ++t){
if (flag[i].test(t) || flag[9 + j].test(t) || flag[18 + (i / 3) * 3 + j / 3].test(t))
continue;

board[i][j] = '1' + t;
flag[i].set(t);
flag[9 + j].set(t);
flag[18 + (i / 3) * 3 + j / 3].set(t);
if (solver(board, m, n, i, j + 1, flag)) return true;
flag[i].set(t, 0);
flag[9 + j].set(t, 0);
flag[18 + (i / 3) * 3 + j / 3].set(t, 0);
board[i][j] = '.';
}

return false;
}
};