题目描述
这是 LeetCode 上的 37. 解数独 ，难度为 困难。
Tag : 「回溯算法」、「DFS」、「数独问题」
编写一个程序，通过填充空格来解决数独问题。
数独的解法需 遵循如下规则：
- 数字 1-9 在每一行只能出现一次。
- 数字 1-9 在每一列只能出现一次。
- 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）
数独部分空格内已填入了数字，空白格用 '.' 表示。
示例：
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输入：board = [["5","3",".",".","7",".",".",".","."], ["6",".",".","1","9","5",".",".","."], [".","9","8",".",".",".",".","6","."], ["8",".",".",".","6",".",".",".","3"], ["4",".",".","8",".","3",".",".","1"], ["7",".",".",".","2",".",".",".","6"], [".","6",".",".",".",".","2","8","."], [".",".",".","4","1","9",".",".","5"], [".",".",".",".","8",".",".","7","9"]] 输出：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"], ["1","9","8","3","4","2","5","6","7"], ["8","5","9","7","6","1","4","2","3"], ["4","2","6","8","5","3","7","9","1"], ["7","1","3","9","2","4","8","5","6"], ["9","6","1","5","3","7","2","8","4"], ["2","8","7","4","1","9","6","3","5"], ["3","4","5","2","8","6","1","7","9"]] 解释：输入的数独如上图所示，唯一有效的解决方案如下所示： 复制代码
提示：
- board.length == 9
- board[i].length == 9
- board[i][j] 是一位数字或者 '.'
- 题目数据 保证 输入数独仅有一个解
回溯解法
和 N 皇后一样，是一道回溯解法裸题。
上一题「36. 有效的数独（中等）」是让我们判断给定的 borad 是否为有效数独。
这题让我们对给定 board 求数独，由于 board 固定是 9*9 的大小，我们可以使用回溯算法去做。
这一类题和 N 皇后一样，属于经典的回溯算法裸题。
这类题都有一个明显的特征，就是数据范围不会很大，如该题限制了范围为 9*9，而 N 皇后的 N 一般不会超过 13。
对每一个需要填入数字的位置进行填入，如果发现填入某个数会导致数独解不下去，则进行回溯。
代码：
class Solution { boolean[][] row = new boolean[9][9]; boolean[][] col = new boolean[9][9]; boolean[][][] cell = new boolean[3][3][9]; public void solveSudoku(char[][] board) { for (int i = 0; i < 9; i++) { for (int j = 0; j < 9; j++) { if (board[i][j] != '.') { int t = board[i][j] - '1'; row[i][t] = col[j][t] = cell[i / 3][j / 3][t] = true; } } } dfs(board, 0, 0); } boolean dfs(char[][] board, int x, int y) { if (y == 9) return dfs(board, x + 1, 0); if (x == 9) return true; if (board[x][y] != '.') return dfs(board, x, y + 1); for (int i = 0; i < 9; i++) { if (!row[x][i] && !col[y][i] && !cell[x / 3][y / 3][i]) { board[x][y] = (char)(i + '1'); row[x][i] = col[y][i] = cell[x / 3][y / 3][i] = true; if (dfs(board, x, y + 1)) { break; } else { board[x][y] = '.'; row[x][i] = col[y][i] = cell[x / 3][y / 3][i] = false; } } } return board[x][y] != '.'; } } 复制代码
- 时间复杂度：在固定
9*9的棋盘里，具有一个枚举方案的最大值（极端情况，假设我们的棋盘刚开始是空的，这时候每一个格子都要枚举，每个格子都有可能从 1 枚举到 9，所以枚举次数为 999 = 729），即复杂度不随数据变化而变化。复杂度为 O(1)O(1)O(1) - 空间复杂度：在固定
9*9的棋盘里，复杂度不随数据变化而变化。复杂度为 O(1)O(1)O(1)
最后
这是我们「刷穿 LeetCode」系列文章的第 No.37 篇，系列开始于 2021/01/01，截止于起始日 LeetCode 上共有 1916 道题目，部分是有锁题，我们将先将所有不带锁的题目刷完。
在这个系列文章里面，除了讲解解题思路以外，还会尽可能给出最为简洁的代码。如果涉及通解还会相应的代码模板。
为了方便各位同学能够电脑上进行调试和提交代码，我建立了相关的仓库：github.com/SharingSour…
在仓库地址里，你可以看到系列文章的题解链接、系列文章的相应代码、LeetCode 原题链接和其他优选题解。
