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:
The problem can be solved by using dp. 
dp[i] = 0 means the number i is not good
dp[i] = 1 means the number is rotated to itself.
dp[i] = 2 means the number is rotated to the different number, so it's a good number. 

Code (Java):

class Solution {
    public int rotatedDigits(int N) {
        int[] dp = new int[N + 1];
        int count = 0;
        for (int i = 0; i <= N; i++) {
            if (i < 10) {
                if (i == 0 || i == 1 || i == 8) {
                    dp[i] = 1;
                } else if (i == 2 || i == 5 || i == 6 || i == 9) {
                    dp[i] = 2;
                    count++;
                }
            } else {
                int a = dp[i / 10];
                int b = dp[i % 10];
                if (a == 1 && b == 1) {
                    dp[i] = 1;
                } else if ((a == 1 && b == 2) || (a == 2 && b >= 1)) {
                    dp[i] = 2;
                    count++;
                }
            }
        }

        return count;
    }
}