1. 题目

2. 解题

• 一个数经过若干次各位数平方和后，会等于它自己
• 使用类似环形链表的快慢指针法，最终快慢指针相遇，若不为1则是不快乐数
  

class Solution {
public:int bitSquareSum(int n) {int sum = 0, bit;while(n > 0){bit = n % 10;sum += bit * bit;n = n / 10;}return sum;}bool isHappy(int n) {int slow = n, fast = n;do{slow = bitSquareSum(slow);fast = bitSquareSum(fast);fast = bitSquareSum(fast);}while(slow != fast);return slow == 1;}
};