Leetcode: Valid Sudoku

验证数独有效性

最新推荐文章于 2019-08-14 07:19:46 发布

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最新推荐文章于 2019-08-14 07:19:46 发布

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本文介绍了一种高效的方法来验证数独的有效性，包括检查部分填充的数独板，并且提供了两个实现版本，一个适用于简单的解法，另一个适用于复杂场景。

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 '.'.

网上看到的最优做法：

 1 public class Solution {
 2 public boolean isValidSudoku(char[][] board) {
 3     for (int i=0; i<9; i++) {
 4         if (!isParticallyValid(board,i,0,i,8)) return false;
 5         if (!isParticallyValid(board,0,i,8,i)) return false;
 6     }
 7     for (int i=0;i<3;i++){
 8         for(int j=0;j<3;j++){
 9             if (!isParticallyValid(board,i*3,j*3,i*3+2,j*3+2)) return false;
10         }
11     }
12     return true;
13 }
14 private boolean isParticallyValid(char[][] board, int x1, int y1,int x2,int y2){
15     Set singleSet = new HashSet();
16     for (int i= x1; i<=x2; i++){
17         for (int j=y1;j<=y2; j++){
18             if (board[i][j]!='.') if(!singleSet.add(board[i][j])) return false;
19         }
20     }
21     return true;
22 }
23 }

然后在解Sudoku Solver的时候，遇到了一个很简单的解法(但是这里不适用)：

 1 public class Solution {
 2     public boolean isValidSudoku(char[][] board) {
 3         if (board == null || board.length != 9 || board[0].length != 9) return false;
 4         for (int i = 0; i < 9; i++) {
 5             for (int j = 0; j < 9; j++) {
 6                 if (board[i][j] == '.') continue;
 7                 if (!isvalid(board, i, j)) return false;
 8             }
 9         }
10         return true;
11     }
12     
13     public boolean isvalid(char[][] board, int i, int j) {
14         for (int a = 0; a < 9; a++) {
15             if (a != i && board[a][j] == board[i][j]) return false;
16         }
17         
18         for (int b = 0; b < 9; b++) {
19             if (b != j && board[i][b] == board[i][j]) return false; 
20         }
21         
22         for (int c = i/3*3; c < i/3*3 + 3; c++) {
23             for (int d = j/3*3; d < j/3*3 + 3; d++) {
24                 if ((c != i || d != j) && board[c][d] == board[i][j]) return false;
25             }
26         }
27         
28         return true;
29     }
30 }