22.1.9.1

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。

public boolean isValidSudoku(char board[][]) {
        int length = board.length;
        //二维数组line表示的是对应的行中是否有对应的数字，比如line[0][3]
        //表示的是第0行（实际上是第1行，因为数组的下标是从0开始的）是否有数字3
        int line[][] = new int[length][length];
        int column[][] = new int[length][length];
        int cell[][] = new int[length][length];
        for (int i = 0; i < length; ++i)
            for (int j = 0; j < length; ++j) {
                //如果还没有填数字，直接跳过
                if (board[i][j] == '.')
                    continue;
                //num是当前格子的数字
                int num = board[i][j] - '0' - 1;
                //k是第几个单元格，9宫格数独横着和竖着都是3个单元格
                int k = i / 3 * 3 + j / 3;
                //如果当前数字对应的行和列以及单元格，只要一个由数字，说明冲突了，直接返回false。
                //举个例子，如果line[i][num]不等于0，说明第i（i从0开始）行有num这个数字。
                if (line[i][num] != 0 || column[j][num] != 0 || cell[k][num] != 0)
                    return false;
                //表示第i行有num这个数字，第j列有num这个数字，对应的单元格内也有num这个数字
                line[i][num] = column[j][num] = cell[k][num] = 1;
            }
        return true;
    }

作者：数据结构和算法
链接：https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x2f9gg/?discussion=LKIXWG
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。

挺有意思的。确实想不到这种办法。

public boolean isValidSudoku(char[][] board) {
        int[] line = new int[9];
        int[] column = new int[9];
        int[] cell = new int[9];
        int shift = 0;
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                //如果还没有填数字，直接跳过
                if (board[i][j] == '.')
                    continue;
                shift = 1 << (board[i][j] - '0');
                int k = (i / 3) * 3 + j / 3;
                //如果对应的位置只要有一个大于0，说明有冲突，直接返回false
                if ((column[i] & shift) > 0 || (line[j] & shift) > 0 || (cell[k] & shift) > 0)
                    return false;
                column[i] |= shift;
                line[j] |= shift;
                cell[k] |= shift;
            }
        }
        return true;
    }

作者：数据结构和算法
链接：https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x2f9gg/?discussion=LKIXWG
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。

 

标签：num,数字,int,length,22.1,board,line,9.1
来源： https://www.cnblogs.com/dark-blue/p/15782411.html