题目：
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:
解法：
1. 利用set保存每次计算的结果，若出现1，则返回true，若出现已有元素，而没有出现1则说明会陷入循环，返回false
class Solution { public boolean isHappy(int n) { Set<Integer> set = new HashSet<>(); int newn = 0; while(!set.contains(n)) { set.add(n); newn = 0; while(n>0) { newn += (n%10)*(n%10); n /= 10; } if(newn==1) return true; n = newn; } return false; } }
2. 参考了discussion的做法，经规律总结，当计算结果在10以内时，只有7和1才会是happy number
class Solution { public boolean isHappy(int n) { int newn = 0; while(n>9) { newn = 0; while(n>0) { newn += (n%10)*(n%10); n /= 10; } n = newn; } return n==7||n==1; } }
3. 由于计算结果最终会形成一个循环，可以参照Floyd判圈算法的思想，设置一个slow一次计算一步，一个fast一次计算两步，则当走入循环，fast会在一圈以内追上slow，使得两者此时指向的结果相同
class Solution { public boolean isHappy(int n) { int slow = n; int fast = n; do{ slow = helper(slow); fast = helper(helper(fast)); if(slow==1||fast==1) return true; }while(slow!=fast); return false; } public int helper(int n) { int res = 0; while(n>0) { res += (n%10)*(n%10); n /= 10; } return res; } }
原文：https://www.cnblogs.com/trymorel/p/12622790.html
