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: 

Input: 19
Output: true
Explanation: 
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1

题目大意：

 　　给定正整数，按照示例方式用它每个数的平方和替代它，若平方和等于1则该正整数为快乐数；若一直无限循环则非快乐数。

理  解：

 　　对于正整数n，n%10从后往前遍历它每一个数，n = n/10删除已遍历的数，sum表示平方和，若sum=1，则n为快乐数；否则，n = sum重新计算平方和。

　　用集合set判断是否存在循环，每计算一个数，就保存在set中，若s.count(n)不等于0表明已经计算过该数，出现了循环，n为非快乐数。

代 码 C++：

class Solution {
public:
    bool isHappy(int n) {
        int sum = 0;
        
        set<int> s;
        while(n!=1){
            s.insert(n);
            while(n!=0){
                sum += (n%10)*(n%10);
                n = n/10;
            }
        
            n = sum;
            if(s.count(n)!=0)
                return false;
            sum = 0;
        }
        return true;
    }
};

运行结果：

　　执行用时 :4 ms, 在所有C++提交中击败了97.12%的用户

　　内存消耗 :8.5 MB, 在所有C++提交中击败了11.41%的用户