Leetcode 788. Rotated Digits

X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X.  Each digit must be rotated - we cannot choose to leave it alone.

A number is valid if each digit remains a digit after rotation. 0, 1, and 8 rotate to themselves; 2 and 5 rotate to each other; 6 and 9 rotate to each other, and the rest of the numbers do not rotate to any other number and become invalid.

Now given a positive number N, how many numbers X from 1 to N are good?

Example:
Input: 10
Output: 4
Explanation: 
There are four good numbers in the range [1, 10] : 2, 5, 6, 9.
Note that 1 and 10 are not good numbers, since they remain unchanged after rotating.

Note:

• N  will be in range [1, 10000].

Solution:

Brute-force. Check each number from 1 to N and see if it's a good number.

Code (Java):

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class Solution {     public int rotatedDigits(int N) {         int count = 0;                   for (int num = 1; num <= N; num++) {             if (isGoodNumber(num)) {                 count++;             }         }                   return count;     }           private boolean isGoodNumber(int num) {         boolean has2569 = false;         while (num > 0) {             int digit = num % 10;             has2569 |= contains2569(digit);                           if (contains347(digit)) {                 return false;             }             num /= 10;         }                   return has2569;     }           private boolean contains347(int digit) {         if (digit == 3 || digit == 4 || digit == 7) {             return true;         }                   return false;     }           private boolean contains2569(int digit) {         if (digit == 2 || digit == 5 || digit == 6 || digit == 9) {             return true;         }                   return false;     } }