解數(shù)獨(dú)

• https://leetcode.cn/problems/sudoku-solver/description/

描述

• 編寫(xiě)一個(gè)程序，通過(guò)填充空格來(lái)解決數(shù)獨(dú)問(wèn)題
• 數(shù)獨(dú)的解法需 遵循如下規(guī)則：
  • 數(shù)字 1-9 在每一行只能出現(xiàn)一次
  • 數(shù)字 1-9 在每一列只能出現(xiàn)一次
  • 數(shù)字 1-9 在每一個(gè)以粗實(shí)線分隔的 3x3 宮內(nèi)只能出現(xiàn)一次。（請(qǐng)參考示例圖）
• 數(shù)獨(dú)部分空格內(nèi)已填入了數(shù)字，空白格用 ‘.’ 表示

示例 1

輸入：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"]]
輸出：[["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"]]

解釋：輸入的數(shù)獨(dú)如上圖所示，唯一有效的解決方案如下所示：

提示

• board.length == 9
• board[i].length == 9
• board[i][j] 是一位數(shù)字或者 ‘.’
• 題目數(shù)據(jù) 保證 輸入數(shù)獨(dú)僅有一個(gè)解

Typescript 版算法實(shí)現(xiàn)

1 ) 方案1: 回溯

/**Do not return anything, modify board in-place instead.*/
function solveSudoku(board: string[][]): void {const line: boolean[][] = Array.from({ length: 9 }, () => Array(9).fill(false));const column: boolean[][] = Array.from({ length: 9 }, () => Array(9).fill(false));const block: boolean[][][] = Array.from({ length: 3 }, () => Array.from({ length: 3 }, () => Array(9).fill(false)));const spaces: number[][] = [];// 初始化狀態(tài)for (let i = 0; i < 9; i++) {for (let j = 0; j < 9; j++) {if (board[i][j] === '.') {spaces.push([i, j]);} else {const digit = parseInt(board[i][j]) - 1;line[i][digit] = true;column[j][digit] = true;block[Math.floor(i / 3)][Math.floor(j / 3)][digit] = true;}}}// 深度優(yōu)先搜索函數(shù)function dfs(pos: number): boolean {if (pos === spaces.length) {return true;}const [i, j] = spaces[pos];for (let digit = 0; digit < 9; digit++) {if (!line[i][digit] && !column[j][digit] && !block[Math.floor(i / 3)][Math.floor(j / 3)][digit]) {line[i][digit] = true;column[j][digit] = true;block[Math.floor(i / 3)][Math.floor(j / 3)][digit] = true;board[i][j] = (digit + 1).toString();if (dfs(pos + 1)) {return true;}// 回溯line[i][digit] = false;column[j][digit] = false;block[Math.floor(i / 3)][Math.floor(j / 3)][digit] = false;board[i][j] = '.';}}return false;}dfs(0);
}

2 ) 方案2: 位運(yùn)算優(yōu)化

/**Do not return anything, modify board in-place instead.*/
function solveSudoku(board: string[][]): void {const line: number[] = Array(9).fill(0);const column: number[] = Array(9).fill(0);const block: number[][] = Array.from({ length: 3 }, () => Array(3).fill(0));const spaces: number[][] = [];// 翻轉(zhuǎn)函數(shù)，用于設(shè)置或清除位function flip(i: number, j: number, digit: number) {const mask = 1 << digit;line[i] ^= mask;column[j] ^= mask;block[Math.floor(i / 3)][Math.floor(j / 3)] ^= mask;}// 初始化狀態(tài)for (let i = 0; i < 9; i++) {for (let j = 0; j < 9; j++) {if (board[i][j] === '.') {spaces.push([i, j]);} else {const digit = parseInt(board[i][j]) - 1;flip(i, j, digit);}}}// 深度優(yōu)先搜索函數(shù)function dfs(pos: number): boolean {if (pos === spaces.length) {return true;}const [i, j] = spaces[pos];let mask = 0x1ff & ~(line[i] | column[j] | block[Math.floor(i / 3)][Math.floor(j / 3)]);while (mask > 0) {const digit = Math.log2(mask & -mask); // 找到最右側(cè)的 1 的位置flip(i, j, digit);board[i][j] = (digit + 1).toString();if (dfs(pos + 1)) {return true;}flip(i, j, digit);mask &= mask - 1; // 將最右側(cè)的 1 置為 0}return false;}dfs(0);
}

3 ) 方案3: 枚舉優(yōu)化

/**Do not return anything, modify board in-place instead.*/
function solveSudoku(board: string[][]): void {const line: number[] = Array(9).fill(0);const column: number[] = Array(9).fill(0);const block: number[][] = Array.from({ length: 3 }, () => Array(3).fill(0));const spaces: number[][] = [];// 翻轉(zhuǎn)函數(shù)，用于設(shè)置或清除位function flip(i: number, j: number, digit: number) {const mask = 1 << digit;line[i] ^= mask;column[j] ^= mask;block[Math.floor(i / 3)][Math.floor(j / 3)] ^= mask;}// 初始化已知數(shù)字的狀態(tài)for (let i = 0; i < 9; i++) {for (let j = 0; j < 9; j++) {if (board[i][j] !== '.') {const digit = parseInt(board[i][j]) - 1;flip(i, j, digit);}}}// 嘗試通過(guò)唯一解填充空格let modified: boolean;do {modified = false;for (let i = 0; i < 9; i++) {for (let j = 0; j < 9; j++) {if (board[i][j] !== '.') {continue;}let mask = 0x1ff & ~(line[i] | column[j] | block[Math.floor(i / 3)][Math.floor(j / 3)]);if ((mask & (mask - 1)) === 0 && mask !== 0) { // mask 的二進(jìn)制表示僅有一個(gè) 1const digit = Math.log2(mask); // 找到最右側(cè)的 1 的位置flip(i, j, digit);board[i][j] = (digit + 1).toString();modified = true;}}}} while (modified);// 記錄剩余未填充的空格for (let i = 0; i < 9; i++) {for (let j = 0; j < 9; j++) {if (board[i][j] === '.') {spaces.push([i, j]);}}}// 深度優(yōu)先搜索函數(shù)function dfs(pos: number): boolean {if (pos === spaces.length) {return true;}const [i, j] = spaces[pos];let mask = 0x1ff & ~(line[i] | column[j] | block[Math.floor(i / 3)][Math.floor(j / 3)]);while (mask > 0) {const digit = Math.log2(mask & -mask); // 找到最右側(cè)的 1 的位置flip(i, j, digit);board[i][j] = (digit + 1).toString();if (dfs(pos + 1)) {return true;}flip(i, j, digit);mask &= mask - 1; // 將最右側(cè)的 1 置為 0}return false;}dfs(0);
}