Leetcode: Happy Number

Write an algorithm to determine if a number is "happy".

A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.

Example: 19 is a happy number

1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1

First Time: Not good, when I try to get to the most significant bit, I define current bit called cur, and cur *= 10 each time. This might gives rise to overflow in line 12 when cur * 10. To overcome this I have to define cur to be a long.

The best way to get each digit is to use n%10, and n /=10 each loop. Dividing will avoid unnecessary overflow.

BTW, no need to define an arraylist to store all the digits.

public class Solution {
    public boolean isHappy(int n) {
        if (n <= 0) return false;
        if (n == 1) return true;
        HashSet<Integer> set = new HashSet<Integer>();
        while(!set.contains(n)) {
            set.add(n);
            long cur = 1;
            ArrayList<Integer> digits = new ArrayList<Integer>();
            int sum = 0;
            while (n / cur >= 1) {
                digits.add((int)(n % (cur*10) / cur));
                cur *= 10;
            }
            for (int digit : digits) {
                sum += (int)Math.pow(digit, 2);
            }
            n = sum;
            if (n == 1) return true;
        }
        return false;
    }
}

About syntax, line 16 it is correct to write it as: sum += Math.pow(digit, 2),

but it is incorrect as: sum = sum + Math.pow(digit, 2), should write as sum = sum + (int)Math.pow(digit, 2)

So better cast pow into integer before use it.

 

Second Time: Better

The best way to get each digit is to use n%10, and n /=10 each loop. 

public class Solution {
    public boolean isHappy(int n) {
        if (n < 0) return false;
        if (n == 1) return true;
        HashSet<Integer> set = new HashSet<Integer>();
        while (n != 1) {
            if (set.contains(n)) return false;
            set.add(n);
            int sum = 0;
            while (n > 0) {
                sum += (int)Math.pow(n%10, 2);
                n /= 10;
            }
            n = sum;
        }
        return true;
    }
}