Solution in Java using HashSet

  • 0
    public boolean isHappy(int n) {
        HashSet<Integer> set = new HashSet();
        return helper(n, set);
    public boolean helper(int n, HashSet<Integer> set){
        if(n == 1)
            return true;
        int num = 0;
        while(n > 0){
            num += Math.pow(n % 10, 2);
            n /= 10;
            return helper(num, set);
        return false;

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    Nice and clear, but it does not consume constant space on the stack; i.e. helper is not tail recursive. i believe that an iterative solution in this particular case is also more elegant.

  • 0

    Thank you for your reply. :) You are right, it is not a constant space solution. I did some research on tail recursive and realised that Java might not support tail recursive optimization though.

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