leetcode 37. 解数独
leetcode 37. 解数独
编写一个程序，通过填充空格来解决数独问题。
数独的解法需 遵循如下规则：
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）
数独部分空格内已填入了数字，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：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：
提示：
board.length == 9
board[i].length == 9
board[i][j] 是一位数字或者 '.'
题目数据 保证 输入数独仅有一个解
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/sudoku-solver
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。
class Solution { public void solveSudoku(char[][] matrix) {
bb(matrix, 0, 0);
} public boolean bb(char[][] matrix, int x, int y) { if (x == 8 && y == 9) { return true;
} if (y == 9) {
y = 0;
x++;
} if (matrix[x][y] != '.') { return bb(matrix, x, y + 1);
} boolean flag = false;
List<Character> cc = cc(matrix, x, y); if (cc.size() == 0) { return flag;
} for (Character cter : cc) {
matrix[x][y] = cter; if (bb(matrix, x, y + 1)) { return true;
} else {
matrix[x][y] = '.';
}
} return flag;
} public List<Character> cc(char[][] matrix, int x, int y) {
List<Character> list = new ArrayList<>(8); byte[] arr = new byte[9]; for (int i = 0; i < 9; i++) { if (matrix[x][i] != '.') {
arr[matrix[x][i] - 49] = 1;
} if (matrix[i][y] != '.') {
arr[matrix[i][y] - 49] = 1;
}
} int i = x / 3 * 3; int j = y / 3 * 3; for (int k = 0; k < 3; k++) { for (int m = 0; m < 3; m++) { if (matrix[k + i][m + j] != '.') {
arr[matrix[k + i][m + j] - 49] = 1;
}
}
} for (int k = 0; k < 9; k++) { if (arr[k] != 1) {
list.add((char) (k + 49));
}
} return list;
}
}
