题目出处
题目描述
个人解法
思路：
java
todo代码示例：（Java）
java
todo复杂度分析
java
todo官方解法
方法1：回溯
思路：
代码示例：（Java）
java
public class Solution1 {
private boolean[][] line = new boolean[9][9];
private boolean[][] column = new boolean[9][9];
private boolean[][][] block = new boolean[3][3][9];
private boolean valid = false;
private List<int[]> spaces = new ArrayList<int[]>();
public void solveSudoku(char[][] board) {
for (int i = 0; i < 9; ++i) {
for (int j = 0; j < 9; ++j) {
if (board[i][j] == '.') {
spaces.add(new int[]{i, j});
} else {
int digit = board[i][j] - '0' - 1;
line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = true;
}
}
}
dfs(board, 0);
}
public void dfs(char[][] board, int pos) {
if (pos == spaces.size()) {
valid = true;
return;
}
int[] space = spaces.get(pos);
int i = space[0], j = space[1];
for (int digit = 0; digit < 9 && !valid; ++digit) {
if (!line[i][digit] && !column[j][digit] && !block[i / 3][j / 3][digit]) {
line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = true;
board[i][j] = (char) (digit + '0' + 1);
dfs(board, pos + 1);
line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = false;
}
}
}
}复杂度分析
- 时间复杂度：O(1)。数独共有 81 个单元格，只需要对每个单元格遍历一次即可。
- 空间复杂度：O(1)。由于数独的大小固定，因此哈希表的空间也是固定的。
方法2:位运算优化
思路：
代码示例：（Java）
java
public class Solution2 {
private int[] line = new int[9];
private int[] column = new int[9];
private int[][] block = new int[3][3];
private boolean valid = false;
private List<int[]> spaces = new ArrayList<int[]>();
public void solveSudoku(char[][] board) {
for (int i = 0; i < 9; ++i) {
for (int j = 0; j < 9; ++j) {
if (board[i][j] == '.') {
spaces.add(new int[]{i, j});
} else {
int digit = board[i][j] - '0' - 1;
flip(i, j, digit);
}
}
}
dfs(board, 0);
}
public void dfs(char[][] board, int pos) {
if (pos == spaces.size()) {
valid = true;
return;
}
int[] space = spaces.get(pos);
int i = space[0], j = space[1];
int mask = ~(line[i] | column[j] | block[i / 3][j / 3]) & 0x1ff;
for (; mask != 0 && !valid; mask &= (mask - 1)) {
int digitMask = mask & (-mask);
int digit = Integer.bitCount(digitMask - 1);
flip(i, j, digit);
board[i][j] = (char) (digit + '0' + 1);
dfs(board, pos + 1);
flip(i, j, digit);
}
}
public void flip(int i, int j, int digit) {
line[i] ^= (1 << digit);
column[j] ^= (1 << digit);
block[i / 3][j / 3] ^= (1 << digit);
}
}复杂度分析
- 时间复杂度：O(1)。数独共有 81 个单元格，只需要对每个单元格遍历一次即可。
- 空间复杂度：O(1)。由于数独的大小固定，因此哈希表的空间也是固定的。
方法3:枚举优化
思路：
代码示例：（Java）
java
public class Solution3 {
private int[] line = new int[9];
private int[] column = new int[9];
private int[][] block = new int[3][3];
private boolean valid = false;
private List<int[]> spaces = new ArrayList<int[]>();
public void solveSudoku(char[][] board) {
for (int i = 0; i < 9; ++i) {
for (int j = 0; j < 9; ++j) {
if (board[i][j] != '.') {
int digit = board[i][j] - '0' - 1;
flip(i, j, digit);
}
}
}
while (true) {
boolean modified = false;
for (int i = 0; i < 9; ++i) {
for (int j = 0; j < 9; ++j) {
if (board[i][j] == '.') {
int mask = ~(line[i] | column[j] | block[i / 3][j / 3]) & 0x1ff;
if ((mask & (mask - 1)) == 0) {
int digit = Integer.bitCount(mask - 1);
flip(i, j, digit);
board[i][j] = (char) (digit + '0' + 1);
modified = true;
}
}
}
}
if (!modified) {
break;
}
}
for (int i = 0; i < 9; ++i) {
for (int j = 0; j < 9; ++j) {
if (board[i][j] == '.') {
spaces.add(new int[]{i, j});
}
}
}
dfs(board, 0);
}
public void dfs(char[][] board, int pos) {
if (pos == spaces.size()) {
valid = true;
return;
}
int[] space = spaces.get(pos);
int i = space[0], j = space[1];
int mask = ~(line[i] | column[j] | block[i / 3][j / 3]) & 0x1ff;
for (; mask != 0 && !valid; mask &= (mask - 1)) {
int digitMask = mask & (-mask);
int digit = Integer.bitCount(digitMask - 1);
flip(i, j, digit);
board[i][j] = (char) (digit + '0' + 1);
dfs(board, pos + 1);
flip(i, j, digit);
}
}
public void flip(int i, int j, int digit) {
line[i] ^= (1 << digit);
column[j] ^= (1 << digit);
block[i / 3][j / 3] ^= (1 << digit);
}
}复杂度分析
- 时间复杂度：O(1)。数独共有 81 个单元格，只需要对每个单元格遍历一次即可。
- 空间复杂度：O(1)。由于数独的大小固定，因此哈希表的空间也是固定的。
考察知识点
收获
1.位运算
