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【Leecode】Leecode刷题之路第37天之解数独

news 2024/11/2 0:23:35

题目出处

37-解数独-题目出处

题目描述

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个人解法

思路：

todo

代码示例：（Java）

todo

复杂度分析

todo

官方解法

37-解数独-官方解法

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方法1：回溯

思路：

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代码示例：（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：

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代码示例：（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：

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代码示例：（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);}}