Leetcode-Sudoku Solver

LiBlog發表於2014-12-22
LeetCode

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...

Analysis:

Use three list of HashSet<Integer> to indicate the current availability of each row, col and block. Use recursive method. At each point, iterate through every available number, take the number out from the avaialble sets of the row, col and block, and then go to next point.

Solution:

 1 public class Solution {
 2     public void solveSudoku(char[][] board) {
 3         if (board.length!=9) return;
 4         if (board[0].length!=9) return;
 5 
 6         //Construct the available set for each point.
 7         List<Set<Integer>> rowAva = new ArrayList<Set<Integer>>();
 8         List<Set<Integer>> colAva = new ArrayList<Set<Integer>>();
 9         List<Set<Integer>> areaAva = new ArrayList<Set<Integer>>();
10         Set<Integer> base = new HashSet<Integer>();
11         for (int i=1;i<=9;i++) base.add(i);
12         for (int i=0;i<9;i++){
13             Set<Integer> temp = new HashSet<Integer>();
14             temp.addAll(base);
15             rowAva.add(temp);
16             temp = new HashSet<Integer>();
17             temp.addAll(base);
18             colAva.add(temp);
19             temp = new HashSet<Integer>();
20             temp.addAll(base);
21             areaAva.add(temp);
22         }
23         //Preprocessing: take out each exsiting number from the ava set of the row and the col.
24         for (int i=0;i<9;i++)
25             for (int j=0;j<9;j++)
26                 if (board[i][j]!='.'){
27                     int val = board[i][j]-'0';
28                     rowAva.get(i).remove(val);
29                     colAva.get(j).remove(val);
30                     int ind = (i/3)*3+(j/3);
31                     areaAva.get(ind).remove(val);
32                 }
33     
34         //Recursively find out the solution.
35         solveSudokuRecur(board,0,0,rowAva,colAva,areaAva);
36     }
37 
38     public boolean solveSudokuRecur(char[][] board, int x, int y, List<Set<Integer>> rowAva, List<Set<Integer>> colAva, List<Set<Integer>> areaAva){
39         if (x==9) return true;
40 
41         char cur = board[x][y];
42         int nextX = (x*9+y+1)/9;
43         int nextY = (x*9+y+1)%9;
44         //it is filled.
45         if (cur!='.')  return solveSudokuRecur(board,nextX,nextY,rowAva,colAva,areaAva);
46         //if it is not filled, then try to fill every possible value.
47         int ind = (x/3)*3+(y/3);
48         for (int i=1;i<=9;i++)
49             if (rowAva.get(x).contains(i) && colAva.get(y).contains(i) && areaAva.get(ind).contains(i)){
50                 board[x][y] = (char)(i+'0');
51                 //remove i from ava sets.
52                 rowAva.get(x).remove(i);
53                 colAva.get(y).remove(i);
54                 areaAva.get(ind).remove(i);                
55                 boolean valid = solveSudokuRecur(board,nextX,nextY,rowAva,colAva,areaAva);
56                 if (valid) return true;
57                 //backtracking.
58                 areaAva.get(ind).add(i);
59                 colAva.get(y).add(i);
60                 rowAva.get(x).add(i);
61                 board[x][y]='.';
62             }
63         return false;
64     }
65             
66             
67 }

 

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