【Leetcode】Valid Sudoku

题目链接：https://leetcode.com/problems/valid-sudoku/

题目：

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character ‘.‘.

A partially filled sudoku which is valid.

Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

思路：

每行、每列、每个黑粗线内小方块的数分别用HashSet保存判断是否唯一。

算法：

[java] view plain copy

 

1. </pre><pre name="code" class="java">    public boolean isValidSudoku(char[][] board) {  
2.         Set<Character> row = new HashSet<Character>();  
3.         Set<Character> column = new HashSet<Character>();  
4.         Set<Character> subBox = new HashSet<Character>();  
5.   
6.         for (int i = 0; i < board.length; i++) {  
7.             for (int j = 0; j < board[0].length; j++) {  
8.                 // System.out.print(board[i][j]+" ");  
9.                 if (board[i][j] != ‘.‘) {  
10.                     if (column.contains(board[i][j])) {  
11.                         return false;  
12.                     } else {  
13.                         column.add(board[i][j]);  
14.                     }  
15.                 }  
16.                 // System.out.println();  
17.             }  
18.             column.clear();  
19.         }  
20.   
21.         // System.out.println("=============");  
22.         for (int i = 0; i < board.length; i++) {  
23.             for (int j = 0; j < board[0].length; j++) {  
24.                 if (board[j][i] != ‘.‘) {  
25.                     if (row.contains(board[j][i])) {  
26.                         return false;  
27.                     } else {  
28.                         row.add(board[j][i]);  
29.                     }  
30.                 }  
31.             }  
32.             row.clear();  
33.         }  
34.   
35.         for (int i = 0; i < board.length; i += 3) {  
36.             for (int j = 0; j < board[0].length; j += 3) {  
37.                 for (int ii = i; ii < i + 3; ii++) {  
38.                     for (int jj = j; jj < j + 3; jj++) {  
39.                         if (board[ii][jj] != ‘.‘) {  
40.                             if (subBox.contains(board[ii][jj])) {  
41.                                 return false;  
42.                             } else {  
43.                                 subBox.add(board[ii][jj]);  
44.                             }  
45.                         }  
46.                     }  
47.                 }  
48.                 subBox.clear();  
49.             }  
50.         }  
51.         return true;  
52.     }  

【Leetcode】Valid Sudoku

原文：http://blog.csdn.net/yeqiuzs/article/details/51628530