【leetcode刷题笔记】Sudoku Solver

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle...

...and its solution numbers marked in red.

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题解：递归。在每个空位上尝试放置0~9的数，然后递归的解决剩下的空位。

单独写一个判断board目前(x,y)处的数是否合法的函数 public boolean isValidSudoku(char[][] board,int x,int y) ，它就只用检查跟(x,y)同行，同列和同一个九宫格的元素是否和board[x][y]有重复即可。（其实在九宫格中只用判断4个和(x,y)不同行列的元素，因为和(x,y)同行列的我们已经判断过了）。

代码如下：

 1 public class Solution {

 2        public boolean isValidSudoku(char[][] board,int x,int y) {       

 3         //check for row x

 4         for(int i = 0;i < 9;i++)

 5             if(i!=y && board[x][i] == board[x][y])

 6                 return false;

 7         

 8         //check for column y

 9         for(int i = 0;i < 9;i++)

10             if(i!= x &&board[i][y] == board[x][y])

11                 return false;

12         

13         //check for the 3*3 square (x,y) belongs to

14         for(int i = 3 * (x/3);i<3*(x/3)+3;i++){

15             for(int j = 3*(y/3);j<3*(y/3)+3;j++){

16                 if(i!=x && j != y && board[i][j] == board[x][y] )

17                     return false;

18             }

19         }

20         

21         return true;

22     }

23     

24     private boolean solveSudokuRecur(char[][] board){

25         for(int i = 0;i < 9;i++){

26             for(int j = 0;j < 9;j++){

27                 if(board[i][j] != '.')

28                     continue;

29                 for(int k = 1;k <= 9;k++){

30                     board[i][j] = (char)(k + '0');

31                     if(isValidSudoku(board,i,j) && solveSudokuRecur(board))

32                         return true;

33                     board[i][j] = '.';

34                 }

35                 return false;

36             }

37         }

38         return true;

39     }

40     public void solveSudoku(char[][] board) {

41         solveSudokuRecur(board);

42     }

43 }

注意之前做过的Valid Sudoku这道题是判断整个数独是否合法，而不是单独某个位置(x,y)是否合法，它需要遍历整个数独，所以这段代码不能拿来用了。