LeetCode

题目：

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.

思路：

先验证每一行，再验证每一列，再验证每一个小正方形。

package sudoku;

public class ValidSudoku {

    public boolean isValidSudoku(char[][] board) {
        int n = 9;
        for (int i = 0; i < n; ++i) {
            int[] arr = new int[10];
            for (int j = 0; j < n; ++j) {
                if (board[i][j] != '.')
                    arr[board[i][j] - '0']++;
            }
            if (!validate(arr))
                return false;
        }
        
        for (int i = 0; i < n; ++i) {
            int[] arr = new int[10];
            for (int j = 0; j < n; ++j) {
                if (board[j][i] != '.')
                    arr[board[j][i] - '0']++;
            }
            if (!validate(arr))
                return false;
        }
        
        for (int i = 0; i < n; i += 3) {
            for (int j = 0; j < n; j += 3) {
                int[] arr = new int[10];
                for (int x = i; x < i + 3; ++x) {
                    for (int y = j; y < j + 3; ++y) {
                        if (board[x][y] != '.') 
                            arr[board[x][y] - '0']++;
                    }
                }
                if (!validate(arr))
                    return false;
            }
        }
        
        return true;
    }
    
    private boolean validate(int[] arr) {
        for (int i = 1; i < 10; ++i) {
            if (arr[i] > 1)
                return false;
        }
        return true;
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub

    }

}