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;
	}
};