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.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/happy-number
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这道题的难点我觉得就是，当各个数字的平方和sum不等于1时，不能直接退出。但是继续循环下去的时候又容易进入了一个死循环，因此，我们可以设置一个set来存储出现过的平方和，当新算出来的平方和已经存在于此集合时，证明即将进入死循环，所以可以返回false，如果此平方和未曾出现过时，继续判断，同时把此数添加至set里面，并且把n更新为sum.

代码如下：

class Solution {
    public boolean isHappy(int n) {
        Set<Integer> temp = new HashSet<>();
        while(true)
        {
            int sum = 0;
            while(n != 0)
            {
               int m = n % 10;
                sum += m * m;
                n /= 10;
            }
            if(sum == 1)
            {
                return true;
            }
            else if(temp.contains(sum))
            {
                return false;
            }
            else
            {
                temp.add(sum);
                n= sum;
            }
        }
    }
}