有效的数独----java

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。

数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
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 1：

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
示例 2：

输入：board =
[["8","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：false
解释：除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

提示：

board.length == 9
board[i].length == 9
board[i][j] 是一位数字（1-9）或者 '.'

作者：力扣 (LeetCode)
链接：https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x2f9gg/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

class Solution {
    public boolean isValidSudoku(char board[][]) {
        int len = board.length;     //数组长度
        int[][] row = new int[len][10]; //使用行二维数组，第一个下标代表行，第二个下标代表数字
        int[][] column = new int[len][10];     //使用列二维数组，第一个下标代表行，第二个下标代表数
        int[][] jiugongge = new int[len][10];   //使用九宫格二维数组，第一个下标代表行，第二个下标代表数
        for(int i=0;i<len;i++){     //遍历数组所有元素
            for(int j=0;j<len;j++){
                if(board[i][j]=='.'){   //遇到无元素，跳过
                    continue;
                }
                int num = board[i][j] - '0';    //字符相减就是ascii码相减，53是5，48是0
                int k = j / 3 + (i / 3) * 3;    //j/3是把列九宫格分为3格，i / 3是把行九宫格分为3格，*3是找到在这个小九宫格中的元素
                if(row[i][num]==1||column[j][num]==1||jiugongge[k][num]==1){    //判断行，列，九宫格二维数组中是否存在元素了
                    return false;
                }else{  //不存在元素，我们把内容换1标记
                    row[i][num]=1;
                    column[j][num]=1;
                    jiugongge[k][num]=1;
                }
            }
        }
        return true;
    }
}