36. 有效的数独

判断一个 9x9 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。

9*9

上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字，空白格用 '.' 表示。

示例 1:

输入:
[
["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:
[
["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。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/valid-sudoku
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

从上层往下一层一层判断，难点在于判断当前数字是否合法

class Solution {
    private int targetN = 9;// 大宫格的n
    private int blockN = 3; // 小宫格的n
    private boolean[][] columnNumExist; // 列序号-数字:是否存在
    private boolean[][] rowNumExist;    // 行序号-数字：是否存在
    private boolean[][][] blockNumExist;  // 第几个3*3宫-数字：是否存在

    /**
     * 给定数独序列只包含数字 1-9 和字符 '.' 。
     * 给定数独永远是 9x9 形式的。
     * @param board
     * @return
     */
    public boolean isValidSudoku(char[][] board) {
        columnNumExist = new boolean[targetN][targetN];
        rowNumExist = new boolean[targetN][targetN];
        blockNumExist = new boolean[blockN][blockN][targetN];
        for (int i = 0; i < targetN; i++) {
            for (int j = 0; j < targetN; j++) {
                // '.'不需要进行判断
                if ('.' == board[i][j]) {
                    continue;
                }
                // board[i][j] - '0' - 1 => 将字符转化为数字1-9，然后再-1，存储数字0-8
                if (!curValid(i, j, board[i][j] - '0' - 1)) {
                    return false;
                }
                recordNum(i, j, board[i][j] - '0' - 1);
            }
        }
        return true;
    }

    /**
     * 判断当前值在这个坐标上是否满足
     * @param row 行
     * @param column 列
     * @param num 数字
     * @return
     */
    private boolean curValid(int row, int column, int num) {
        // 行是否已经包括
        if (rowNumExist[row][num]) {
            return false;
        }
        // 列是否已经包括
        if (columnNumExist[column][num]) {
            return false;
        }
        // 3*3宫格是否已经包括
        if (blockNumExist[row / blockN][column / blockN][num]) {
            return false;
        }
        return true;
    }

    /**
     * 记录当前坐标的值
     * @param row 行
     * @param column 列
     * @param num 数字
     * @return
     */
    private void recordNum(int row, int column, int num) {
        rowNumExist[row][num] = true;
        columnNumExist[column][num] = true;
        blockNumExist[row / blockN][column / blockN][num] = true;
    }
}

运行效率

进一步优化，用二进制数替代二位数组，二级制的位运算效率很高

class Solution {
    private int targetN = 9;// 大宫格的n
    private int blockN = 3; // 小宫格的n
    private int curRowBit; // 用一个数字表示当前行的情况，相当于 010100000 9个比特位表示，1表示已经选过
    private int[] columnBit;
    private int[] blockBit;

    /**
     * 给定数独序列只包含数字 1-9 和字符 '.' 。
     * 给定数独永远是 9x9 形式的。
     * @param board
     * @return
     */
    public boolean isValidSudoku(char[][] board) {
        columnBit = new int[targetN];
        blockBit = new int[targetN];
        for (int i = 0; i < targetN; i++) {
            curRowBit = 0; // 存储当前行的bit位数
            for (int j = 0; j < targetN; j++) {
                // '.'不需要进行判断
                if ('.' == board[i][j]) {
                    continue;
                }
                // board[i][j] - '0' - 1 => 将字符转化为数字1-9，然后再-1，存储数字0-8
                if (!curValid(i, j, board[i][j] - '1')) {
                    return false;
                }
                recordNum(i, j, board[i][j] - '1');
            }
        }
        return true;
    }

    /**
     * 判断当前值在这个坐标上是否满足
     * @param row 行
     * @param column 列
     * @param num 数字
     * @return
     */
    private boolean curValid(int row, int column, int num) {
        // 行是否已经包括
        if ((curRowBit >> num) % 2 == 1) {
            return false;
        }
        // 列是否已经包括
        if ((columnBit[column] >> num) % 2 == 1) {
            return false;
        }
        // 3*3宫格是否已经包括
        int cnt = row / blockN * blockN + column / blockN;
        if ((blockBit[cnt] >> num) % 2 == 1) {
            return false;
        }
        return true;
    }

    /**
     * 记录当前坐标的值
     * @param row 行
     * @param column 列
     * @param num 数字
     * @return
     */
    private void recordNum(int row, int column, int num) {
        curRowBit += 1 << num;
        columnBit[column] += 1 << num;
        int cnt = row / blockN * blockN + column / blockN;
        blockBit[cnt] += 1 << num;
    }
}

二进制数替代二位数组