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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.
Sudoku规则：
在一个9*9的区域内，
每行1-9出现且只出现一次，
每列1-9出现且只出现一次，
在9个子3*3的区域内1-9出现且只出现一次。
1 public class Solution { 2 public boolean isValidSudoku(char[][] board) { 3 ArrayList<HashSet<Integer>> row = new ArrayList<HashSet<Integer>>(); 4 ArrayList<HashSet<Integer>> col = new ArrayList<HashSet<Integer>>(); 5 ArrayList<HashSet<Integer>> box = new ArrayList<HashSet<Integer>>(); 6 7 for (int i = 0; i < 9; i++) { 8 row.add(new HashSet<Integer>()); 9 col.add(new HashSet<Integer>()); 10 box.add(new HashSet<Integer>()); 11 } 12 13 for (int i = 0; i < 9; i++) { 14 for (int j = 0; j < 9; j++) { 15 if (board[i][j] == ‘.‘) { 16 continue; 17 } 18 if (row.get(i).contains(board[i][j]-‘0‘)) { 19 return false; 20 } else row.get(i).add(board[i][j]-‘0‘); 21 22 if (col.get(j).contains(board[i][j] - ‘0‘)) { 23 return false; 24 }else col.get(j).add(board[i][j] - ‘0‘); 25 26 int index = ((i / 3) * 3) + j / 3; 27 if (box.get(index).contains(board[i][j] - ‘0‘)) { 28 return false; 29 }else box.get(index).add(board[i][j] - ‘0‘); 30 } 31 } 32 33 return true; 34 } 35 }
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原文地址：http://www.cnblogs.com/birdhack/p/4212152.html
