解数独
08/18/2025
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:
数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
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。
示例 :
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"]]
本人题解
/**
* @param {character[][]} board
* @return {void} Do not return anything, modify board in-place instead.
*/
var solveSudoku = function(board) {
const nextCell = getNextCell(board);
if (!nextCell.length) {
return;
}
solve(board, ...nextCell);
console.log(board);
return;
};
function solve(board, x, y) {
if (x === undefined || y === undefined) {
return true;
}
const availableNumbers = getAvailableNumbers(board, x, y);
const nextCell = getNextCell(board, x, y+1);
if (availableNumbers.size === 0) {
return false
}
for (const number of availableNumbers) {
board[x][y] = number;
if (solve(board, ...nextCell)) {
return true
}
}
board[x][y] = '.';
return false;
}
function getNextCell(board, x = 0, y = 0) {
for (let i = x; i < 9; i++) {
for (let j = y; j < 9; j++) {
if (board[i][j] === '.') {
return [i, j];
}
}
y = 0;
}
return [];
}
function getAvailableNumbers(board, x, y) {
const availableNumbers = new Set(["1", "2", "3", "4", "5", "6", "7", "8", "9"]);
const row = board[x];
const col = board.map(row => row[y]);
const square = [];
const squareX = Math.floor(x / 3);
const squareY = Math.floor(y / 3);
for (let i = 0; i < 3; i++) {
for (let j = 0; j < 3; j++) {
square.push(board[squareX * 3 + i][squareY * 3 + j]);
}
}
const numbers = [...row, ...col, ...square];
for (const number of numbers) {
if (number !== '.')
availableNumbers.delete(number);
}
return availableNumbers;
}分析
- 首先要找到下一个要填充的位置
- 找到该位置的 横排 竖列 9宫格,从而找到它 available 的数字
- 尝试每个可用的数字,并且在此基础上,重复 1 2 3
- 如果没有可用的数字,则返回 false,并且回溯。如果每个可用的数字的结果都是false,将值还原为 .,继续回溯
- 当下一个位置是空时,说明已经解决!