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;
        }
    }
}