LeetCode Valid Sudoku

原题链接在这里：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.

题解：

先一行一行检查，再一列一列检查，这里注意外层for loop是j

最后一个一个小方块来检查，关键是如何检查sub box. 这里设值index 0-8 box, loop row with box/3*3 - box/3*3+3, loop column with box%3*3 - box%3*3+3.

Time Complexity: O(n^2). n = board.length. Space: O(1).

AC Java:

 1 public class Solution {
 2     public boolean isValidSudoku(char[][] board) {
 3         if(board == null || board.length != 9 || board[0].length != 9){
 4             return false;
 5         }
 6
 7         HashSet<Character> hs = new HashSet<Character>();
 8         //check each row
 9         for(int i = 0; i<board.length; i++){
10             hs.clear();
11             for(int j = 0; j<board[0].length; j++){
12                 if(board[i][j] != '.'){
13                     if(!hs.contains(board[i][j])){
14                         hs.add(board[i][j]);
15                     }else{
16                         return false;
17                     }
18                 }
19             }
20         }
21
22         //check each column
23         for(int j = 0; j<board[0].length; j++){
24             hs.clear();
25             for(int i = 0; i<board.length; i++){
26                 if(board[i][j] != '.'){
27                     if(!hs.contains(board[i][j])){
28                         hs.add(board[i][j]);
29                     }else{
30                         return false;
31                     }
32                 }
33             }
34         }
35
36         //check each subbox
37         for(int box = 0; box<9; box++){
38             hs.clear();
39             for(int i = box/3*3; i<box/3*3+3; i++){
40                 for(int j = box%3*3; j<box%3*3+3; j++){
41                     if(board[i][j] != '.'){
42                         if(!hs.contains(board[i][j])){
43                             hs.add(board[i][j]);
44                         }else{
45                             return false;
46                         }
47                     }
48                 }
49             }
50         }
51         return true;
52     }
53 }

跟上Sudoku Solver.

转载于:https://www.cnblogs.com/Dylan-Java-NYC/p/4825064.html

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