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 numberN, how many numbers X from1toNare good?

class Solution {
    public int rotatedDigits(int N) {
        int res = 0;
        for(int i = 1; i <= N; i++){
            if(isGoodNumber(i))
                res++;
        }

        return res;
    }

    public boolean isGoodNumber(int N) {
        String n = String.valueOf(N);
        char[] cs = n.toCharArray();
        for(int i = 0; i < cs.length; i++) {
            if(cs[i] == '0' || cs[i] == '1' || cs[i] == '8')
                continue;

            if(cs[i] == '6') {
                cs[i] = '9';
            } else if(cs[i] == '9') {
                cs[i] = '6';
            } else if(cs[i] == '2') {
                cs[i] = '5';
            } else if(cs[i] == '5') {
                cs[i] = '2';
            } else {
                return false;
            }
        }

        return Integer.parseInt(String.valueOf(cs)) != N;
    }
}