【leetcode刷题笔记】Valid Sudoku

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.

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题解：根据题目的意思，每行每列和每一个3*3的九宫格里面1~9这9个数不能有重复的，那么就按行，列，和九宫格一一检查即可，要注意下标的计算和'.'符号的处理。

代码如下：

 1 public class Solution {
 2     public boolean isValidSudoku(char[][] board) {
 3         int length = board.length;
 4         if(length == 0)
 5             return true;
 6         
 7         for(int i = 0;i < length;i++){
 8             boolean[] row_numbers = new boolean[10];
 9             boolean[] column_numbers = new boolean[10];
10             for(int j = 0;j < length;j++){
11                 //check if rows are valid
12                 if(board[i][j]!= '.' ){
13                     if(row_numbers[board[i][j] - '0'])
14                         return false;
15                     row_numbers[board[i][j]-'0'] = true;
16                 }
17                 
18                 //check if colums are valid
19                 if(board[j][i]!= '.'){
20                     if(column_numbers[board[j][i]-'0'])
21                         return false;
22                     column_numbers[board[j][i]-'0'] = true;
23                 }
24             }                
25         }
26         
27         //check if every 3*3 grid is valid
28         for(int i = 0;i < 3;i++){
29             for(int j = 0;j < 3;j++){
30                 boolean[] numbers = new boolean[10];
31                 for(int row = 3*i;row < 3*i+3;row++){
32                     for(int column = 3*j;column < 3*j+3;column++){
33                         if(board[row][column] != '.'){
34                             if(numbers[board[row][column]-'0'])
35                                 return false;
36                             numbers[board[row][column]-'0'] = true;
37                         }
38                     }
39                 }
40             }
41         }
42         
43         return true;
44     }
45 }

代码中行和列的检查在一次9*9的循环中解决了，可以省一点时间，最终耗时532ms。

 

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转载于:https://www.cnblogs.com/sunshineatnoon/p/3858132.html