Leetcode: 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 '.'.

Best Approach:

• '4' in row 7 is encoded as "(4)7".
• '4' in column 7 is encoded as "7(4)".
• '4' in the top-right block is encoded as "0(4)2".

public boolean isValidSudoku(char[][] board) {
    Set seen = new HashSet();
    for (int i=0; i<9; ++i) {
        for (int j=0; j<9; ++j) {
            if (board[i][j] != '.') {
                String b = "(" + board[i][j] + ")";
                if (!seen.add(b + i) || !seen.add(j + b) || !seen.add(i/3 + b + j/3))
                    return false;
            }
        }
    }
    return true;
}

 

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)：

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        if (board == null || board.length != 9 || board[0].length != 9) return false;
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if (board[i][j] == '.') continue;
                if (!isvalid(board, i, j)) return false;
            }
        }
        return true;
    }
    
    public boolean isvalid(char[][] board, int i, int j) {
        for (int a = 0; a < 9; a++) {
            if (a != i && board[a][j] == board[i][j]) return false;
        }
        
        for (int b = 0; b < 9; b++) {
            if (b != j && board[i][b] == board[i][j]) return false; 
        }
        
        for (int c = i/3*3; c < i/3*3 + 3; c++) {
            for (int d = j/3*3; d < j/3*3 + 3; d++) {
                if ((c != i || d != j) && board[c][d] == board[i][j]) return false;
            }
        }
        
        return true;
    }
}