Leetcode 37: 解数独

Leetcode 37: 解数独 是经典的回溯算法问题，考察如何利用递归和剪枝高效求解数独问题。这题主要考察对回溯、递归、深度优先搜索 (DFS)、剪枝优化等算法思想的掌握。

题目描述

给定一个部分填充的数独（9 x 9）网格，用一个有效的算法将其完整解出。

数独规则：
1. 每一行必须包含数字 1-9，不重复。
2. 每一列必须包含数字 1-9，不重复。
3. 每个 3x3 的子盒子必须包含数字 1-9，不重复。

示例输入输出

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"]]

解法 1：回溯法

思路

深度优先搜索：
- 使用 DFS 遍历整个棋盘，每次找到未填的空格，尝试填入 1-9。
- 若当前数字导致冲突（不合法），回溯到上一步重新尝试。
填充判断：
- 判断当前数字是否能加入当前行、列、以及 3x3 小方格中，确保满足数独规则。
终止条件：
- 当所有空格填满时，返回 true。
- 若在当前路径中所有数均无合法选择，则返回 false（触发回溯）。

代码模板

class Solution {public void solveSudoku(char[][] board) {backtrack(board);}private boolean backtrack(char[][] board) {for (int i = 0; i < 9; i++) {for (int j = 0; j < 9; j++) {if (board[i][j] == '.') {for (char num = '1'; num <= '9'; num++) {if (isValid(board, i, j, num)) {board[i][j] = num; // 尝试填入数字if (backtrack(board)) return true;board[i][j] = '.'; // 回溯}}return false; // 如果 1~9 都不行，返回 false，触发回溯}}}return true; // 如果没有剩下的空格，说明填满了}private boolean isValid(char[][] board, int row, int col, char num) {for (int i = 0; i < 9; i++) {// 检查行和列是否出现重复if (board[row][i] == num || board[i][col] == num) return false;// 检查 3x3 小方格是否出现重复int boxRow = 3 * (row / 3) + i / 3;int boxCol = 3 * (col / 3) + i % 3;if (board[boxRow][boxCol] == num) return false;}return true;}
}

复杂度分析

时间复杂度：
- 最坏情况下为 O(9^(n))，其中 n 是未填格子数。每个空格尝试填入 1-9 的数字。
空间复杂度：
- O(n)（递归调用栈的深度）。

适用场景

经典数独问题的首选解法，清晰易实现。
回溯法逻辑直观，可以快速实现并 AC。

解法 2：回溯 + 剪枝（预处理加速）

思路

在基础回溯解法的基础上，添加剪枝优化，通过记录数字的使用情况，避免重复判断，加快搜索速度。

状态记录：
- 使用三个布尔数组分别记录某个数字是否已在当前行、列或 3x3 子盒子 中出现。
- 例如，rows[i][num] 表示数字 num+1 是否出现在第 i 行。
选择联合撤销：
- 每次填一个数字时，更新状态记录；回溯时撤销状态，确保合法性。

代码模板

class Solution {private boolean[][] rows = new boolean[9][9];private boolean[][] cols = new boolean[9][9];private boolean[][] boxes = new boolean[9][9];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] - '1';rows[i][num] = true;cols[j][num] = true;boxes[boxIndex(i, j)][num] = true;}}}backtrack(board, 0, 0);}private boolean backtrack(char[][] board, int row, int col) {// 如果列越界，换到下一行if (col == 9) {col = 0;row++;if (row == 9) return true; // 全部填完}// 如果当前位置已填，跳过if (board[row][col] != '.') {return backtrack(board, row, col + 1);}// 尝试填入数字for (int num = 0; num < 9; num++) {if (!rows[row][num] && !cols[col][num] && !boxes[boxIndex(row, col)][num]) {board[row][col] = (char) ('1' + num);rows[row][num] = true;cols[col][num] = true;boxes[boxIndex(row, col)][num] = true;if (backtrack(board, row, col + 1)) return true;// 回溯撤销选择board[row][col] = '.';rows[row][num] = false;cols[col][num] = false;boxes[boxIndex(row, col)][num] = false;}}return false;}private int boxIndex(int row, int col) {return (row / 3) * 3 + col / 3;}
}