37. 解数独
37. 解数独
🔴 🔖 数组
哈希表
回溯
矩阵
🔗 力扣
LeetCode
题目
Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy all of the following rules :
- Each of the digits
1-9
must occur exactly once in each row. - Each of the digits
1-9
must occur exactly once in each column. - Each of the digits
1-9
must occur exactly once in each of the 93x3
sub-boxes of the grid.
The '.'
character indicates empty cells.
Example 1:
![](https://upload.wikimedia.org/wikipedia/commons/thumb/f/ff/Sudoku- by-L2G-20050714.svg/250px-Sudoku-by-L2G-20050714.svg.png)
Input: 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"]]
Output: [["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"]]
Explanation: The input board is shown above and the only valid solution is shown below:
Constraints:
board.length == 9
board[i].length == 9
board[i][j]
is a digit or'.'
.- It is guaranteed that the input board has only one solution.
题目大意
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。
解题思路
这个问题的核心是使用 回溯法(Backtracking),逐步尝试将每个空格填充数字,并检查当前的填充是否符合数独规则。如果不符合规则,则回退到之前的状态继续尝试其他可能性,直到找到一个可行解。
寻找空格:
- 遍历整个数独,找到尚未填充数字的位置(即空格
' . '
),然后开始尝试填入数字。
- 遍历整个数独,找到尚未填充数字的位置(即空格
递归填充数字:
- 尝试将数字
1-9
依次填入空格中,并在每次填入后,检查行、列、3x3 的小方格是否仍满足数独的规则。
- 尝试将数字
检查有效性:
- 每次填入一个数字后,检查当前数独的有效性(即每行、每列、每个 3x3 小方格中是否包含重复的数字)。
回溯:
- 如果当前数字不满足规则,则回退到上一步,尝试填入下一个可能的数字。
- 如果所有可能的数字都不符合规则,则返回到上一个空格的位置,继续尝试其他可能的数字。
终止条件:
- 当所有空格都填完,并且整个数独符合规则时,递归终止,数独被解出。
复杂度分析
- 时间复杂度:回溯法的时间复杂度非常难以精确估计,因为它取决于输入数独的复杂程度。在最坏情况下,时间复杂度可能会接近
O(9^m)
,其中m
是空格的数量。因为每个空格最多可以填入9
个数字。 - 空间复杂度:
O(1)
,因为数独是在原数组上修改的,且只使用了少量额外空间用于递归栈。
代码
/**
* @param {character[][]} board
* @return {void} Do not return anything, modify board in-place instead.
*/
var solveSudoku = function (board) {
const solve = (board) => {
for (let i = 0; i < 9; i++) {
for (let j = 0; j < 9; j++) {
if (board[i][j] == '.') {
// 尝试填入 1-9
for (let num = 1; num <= 9; num++) {
if (isValid(board, i, j, num + '')) {
board[i][j] = num + '';
// 递归求解
if (solve(board)) {
return true;
}
// 回溯
board[i][j] = '.';
}
}
// 若无法填入任何数字,返回 false
return false;
}
}
}
// 数独已解决
return true;
};
solve(board);
};
// 检查在 board[row][col] 放置 num 是否有效
var isValid = function (board, row, col, num) {
for (let i = 0; i < 9; i++) {
// 检查行和列
if (board[i][col] == num || board[row][i] == num) {
return false;
}
// 检查 3x3 小方格
const boxRow = 3 * ((row / 3) | 0) + ((i / 3) | 0);
const boxCol = 3 * ((col / 3) | 0) + (i % 3);
if (board[boxRow][boxCol] == num) {
return false;
}
}
return true;
};
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