LeetCode第37题_解数独

LeetCode 第37题：解数独

题目描述

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。

难度

困难

题目链接

点击在LeetCode中查看题目

示例

示例 1：

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"]]

提示

• board.length == 9
• board[i].length == 9
• board[i][j] 是一个数字或者 '.'
• 题目数据 保证 输入数独仅有一个解

解题思路

方法：回溯算法

这是一个经典的回溯算法问题，我们需要尝试填充每个空格，并在发现错误时及时回溯。

关键点：

1. 使用回溯法尝试每个可能的数字
2. 优化验证过程，避免重复检查
3. 使用位运算加速判断

具体步骤：

1. 预处理阶段：
   • 记录每行、每列、每个3x3方格中已有的数字
   • 找到所有需要填充的位置
2. 回溯过程：
   • 对每个空位尝试1-9的数字
   • 验证当前数字是否有效
   • 继续填充下一个位置或回溯

时间复杂度：O(9^m)，m为空格数量
空间复杂度：O(m)，递归栈的深度

图解思路

回溯过程分析表

步骤	当前位置	尝试数字	结果	说明
初始状态	(0,2)	1-9	4可用	第一个空格位置
继续填充	(0,3)	1-9	6可用	成功填入4后继续
回溯	(2,0)	1-9	1可用	前面填充正确，继续
完成	(8,8)	-	成功	所有空格已填满

状态记录表

数据结构	用途	实现方式	优势
行状态	记录每行数字	位运算	快速查找和更新
列状态	记录每列数字	位运算	空间效率高
宫状态	记录3x3宫数字	位运算	验证速度快

位运算示例

操作	二进制表示	说明
添加数字5	00010000	第5位置1
删除数字5	11101111	第5位置0
检查数字5	&00010000	与运算判断

代码实现

public class Solution {
    private int[] rows = new int[9];
    private int[] cols = new int[9];
    private int[] boxes = new int[9];
    private bool solved = false;
  
    public void SolveSudoku(char[][] board) {
        // 初始化状态
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if (board[i][j] != '.') {
                    int num = board[i][j] - '0';
                    int pos = 1 << (num - 1);
                    rows[i] |= pos;
                    cols[j] |= pos;
                    boxes[(i/3)*3 + j/3] |= pos;
                }
            }
        }
  
        // 开始回溯
        Backtrack(board, 0, 0);
    }
  
    private bool Backtrack(char[][] board, int row, int col) {
        if (col == 9) {
            row++;
            col = 0;
        }
        if (row == 9) return true;
  
        if (board[row][col] != '.') {
            return Backtrack(board, row, col + 1);
        }
  
        int boxIndex = (row/3)*3 + col/3;
        for (int num = 1; num <= 9; num++) {
            int pos = 1 << (num - 1);
            if ((rows[row] & pos) == 0 && 
                (cols[col] & pos) == 0 && 
                (boxes[boxIndex] & pos) == 0) {
        
                board[row][col] = (char)(num + '0');
                rows[row] |= pos;
                cols[col] |= pos;
                boxes[boxIndex] |= pos;
        
                if (Backtrack(board, row, col + 1)) {
                    return true;
                }
        
                board[row][col] = '.';
                rows[row] &= ~pos;
                cols[col] &= ~pos;
                boxes[boxIndex] &= ~pos;
            }
        }
  
        return false;
    }
}

执行结果

• 执行用时：92 ms
• 内存消耗：38.2 MB

代码亮点

1. 使用位运算优化数字判断
2. 预处理初始状态，避免重复计算
3. 高效的回溯策略
4. 清晰的代码结构和命名

常见错误分析

1. 回溯条件判断错误
2. 状态重置不完整
3. 位运算操作失误
4. 边界条件处理不当

解法对比

解法	时间复杂度	空间复杂度	优点	缺点
暴力回溯	O(9^m)	O(m)	实现简单	效率低
位运算优化	O(9^m)	O(1)	性能好	代码复杂
Dancing Links	O(9^m)	O(n^2)	最优解法	实现难度大

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