Leet Code OJ 202. Happy Number [Difficulty: Easy]
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: 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
翻译：
写一个算法去检测一个数是否是“快乐”的。
“快乐数”按照如下方法定义：首先它是一个正数，使用它的每位数字的平方的和替换它，直到这个数等于1，或者它会一直循环而不到1。这些能够到1终止的数就是“快乐数”。
分析：
本方案采用递归方式去遍历每个数字，直到它小于10，并且预先算出每个10以内的数字定义是否是“快乐数”。
代码：
public class Solution {public boolean isHappy(int n) {if(n>9){int sum=0;int nk=n;while(nk>9){sum+=(nk%10)*(nk%10);nk/=10;}sum+=(nk%10)*(nk%10);return isHappy(sum);}if(n==1||n==7){return true;}else{return false;}}
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