JavaOne Talk: Reversible Numbers

See this post about why this code is being shown. This is a different algorithm, it is a brute force solution to Project Euler problem #145. Here is the Java "reference" implementation.

public class Reversible {
    final int max;
    final int numReversible;

    public Reversible(int max){
        this.max = max;
        this.numReversible = this.countReversible();
    }

    private int countReversible(){
        if (numReversible > 0){
            return numReversible;
        }
        int cnt = 0;
        for (int i=11;i<=max;i++){
            if (reversible(i)) {
                cnt++;
            }
        }
        return cnt;
    }

    private boolean reversible(int n){
        return allOdd(reverse(n));
    }

    private boolean allOdd(int n){
        while (n > 0){
            int remainder = n % 2;
            if (remainder == 0) return false;
            n = n / 10;
        }
        return true;
    }
    
    private int reverse(Integer n){
        char[] digits = n.toString().toCharArray();
        char[] rev = new char[digits.length];
        for (int i=digits.length-1;i>=0;i--){
            rev[i] = digits[digits.length -i-1];
        }
        return n + Integer.parseInt(String.valueOf(rev));
    }
}