[LeetCode] 36. Valid Sudoku

Determine if a 9 x 9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

Each row must contain the digits 1-9 without repetition.
Each column must contain the digits 1-9 without repetition.
Each of the nine 3 x 3 sub-boxes of the grid must contain the digits 1-9 without repetition.

Note:

A Sudoku board (partially filled) could be valid but is not necessarily solvable.
Only the filled cells need to be validated according to the mentioned rules.

Example 1:

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: true

Example 2:

Input: board = 
[["8","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: false
Explanation: Same as Example 1, except with the 5 in the top left corner being modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

Constraints:

board.length == 9
board[i].length == 9
board[i][j] is a digit or '.'.

有效的数独。

请你判断一个 9 x 9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。

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）

注意：

一个有效的数独（部分已被填充）不一定是可解的。
只需要根据以上规则，验证已经填入的数字是否有效即可。
空白格用 '.' 表示。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/valid-sudoku
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这个题没有什么算法，思路是用 hashmap 判断每一行，每一列，和每一个九宫格是否有重复数字，要确保没有重复数字。如果当前的坐标内不是点（.），且当前坐标内的数字存在于 hashmap 则 return false。判断每一行的时候，因为坐标是（i, j），所以判断（j, i）的时候可以同时判断一个对应的列是否满足题意。判断每个 cube 的方式是先要得到 rowIndex 和 colIndex，这里我是当做结论一样背下来的，真的不是很好记。

rowIndex = 3 * (i / 3)
colIndex = 3 * (i % 3)

时间O(mn)

空间O(n)

Java实现

 1 class Solution {
 2     public boolean isValidSudoku(char[][] board) {
 3         boolean[][] rows = new boolean[9][9];
 4         boolean[][] cols = new boolean[9][9];
 5         boolean[][] cubes = new boolean[9][9];
 6         for (int i = 0; i < 9; i++) {
 7             for (int j = 0; j < 9; j++) {
 8                 if (board[i][j] != '.') {
 9                     int num = board[i][j] - '1';
10                     int cubeIndex = i / 3 * 3 + j / 3;
11                     if (rows[i][num] || cols[j][num] || cubes[cubeIndex][num]) {
12                         return false;
13                     } else {
14                         rows[i][num] = true;
15                         cols[j][num] = true;
16                         cubes[cubeIndex][num] = true;
17                     }
18                 }
19             }
20         }
21         return true;
22     }
23 }

JavaScript实现

 1 /**
 2  * @param {character[][]} board
 3  * @return {boolean}
 4  */
 5 var isValidSudoku = function(board) {
 6     let m = board.length;
 7     let n = board[0].length;
 8     for (let i = 0; i < m; i++) {
 9         let rows = new Set();
10         let cols = new Set();
11         let cubes = new Set();
12         for (let j = 0; j < n; j++) {
13             if (board[i][j] != '.') {
14                 if (!rows.has(board[i][j])) {
15                     rows.add(board[i][j]);
16                 } else {
17                     return false;
18                 }
19             }
20             if (board[j][i] != '.') {
21                 if (!cols.has(board[j][i])) {
22                     cols.add(board[j][i]);
23                 } else {
24                     return false;
25                 }
26             }
27             let rowIndex = 3 * Math.floor(i / 3);
28             let colIndex = 3 * (i % 3);
29             if (
30                 board[rowIndex + Math.floor(j / 3)][colIndex + (j % 3)] != '.'
31             ) {
32                 if (
33                     !cubes.has(
34                         board[rowIndex + Math.floor(j / 3)][colIndex + (j % 3)]
35                     )
36                 ) {
37                     cubes.add(
38                         board[rowIndex + Math.floor(j / 3)][colIndex + (j % 3)]
39                     );
40                 } else {
41                     return false;
42                 }
43             }
44         }
45     }
46     return true;
47 };

2021年4月更新

我这里再推荐一个字符串的思路，来自于StefanPochmann大神。这个做法的思想是需要判断某个元素是否是 unique 的，这个思想可以运用到其他需要判断独一性的题目。这里我一时想不起来有哪些题目了，日后有总结我再加以更新。

Java实现

 1 class Solution {
 2     public boolean isValidSudoku(char[][] board) {
 3         HashSet<String> set = new HashSet<>();
 4         for (int i = 0; i < 9; i++) {
 5             for (int j = 0; j < 9; j++) {
 6                 if (board[i][j] != '.') {
 7                     String b = "(" + board[i][j] + ")";
 8                     if (!set.add(b + i) || !set.add(j + b) || !set.add(i / 3 + b + j / 3)) {
 9                         return false;
10                     }
11                 }
12             }
13         }
14         return true;
15     }
16 }

LeetCode 题目总结

[LeetCode] 36. Valid Sudoku

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