【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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