36. 有效的数独

36. 有效的数独

题目描述

请你判断一个 9 x 9 的数独是否有效。只需要 根据以下规则 ，验证已经填入的数字是否有效即可。

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）

注意：

• 一个有效的数独（部分已被填充）不一定是可解的。
• 只需要根据以上规则，验证已经填入的数字是否有效即可。
• 空白格用 '.' 表示。

示例

输入：board = 
[["5","3",".",".","7",".",".",".","."]
,["6",".",".","1","9","5",".",".","."]
,[".","9","8",".",".",".",".","6","."]
,["8",".",".",".","6",".",".",".","3"]
,["4",".",".","8",".","3",".",".","1"]
,["7",".",".",".","2",".",".",".","6"]
,[".","6",".",".",".",".","2","8","."]
,[".",".",".","4","1","9",".",".","5"]
,[".",".",".",".","8",".",".","7","9"]]
输出：true

解题思路

关键在于找到子数独的规律：box_index = (row / 3) * 3 + columns / 3

图片来源于网络，出处不详

class Solution {
    public boolean isValidSudoku(char[][] board) {
        int[][] row = new int[9][9]; //二维数组初始化
        int[][] column = new int[9][9];
        int[][] box = new int[9][9];

        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if (board[i][j] == '.') {
                    continue;
                }
                int num = board[i][j] - '1';//数字1-9对应下标0-8
                int boxIndex = (i/3)*3 + j/3;

                if (row[i][num] > 0 || column[j][num] > 0 || box[boxIndex][num] > 0) {
                    return false;
                }

                row[i][num] = 1;
                column[j][num] = 1;
                box[boxIndex][num] = 1;

            }
        }

        return true;
    }
}