经典数独问题（回溯算法）｜Java

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：

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 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：

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：      

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] != '.';
    }
}      

• 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
• 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

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。