LeetCode 202. Happy Number & 263. Ugly Number

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本文介绍如何通过算法判断一个数是否为快乐数或丑陋数。快乐数是指通过不断将其每位数字的平方和重复替换该数，最终能得到1的正整数；丑陋数则是指那些质因数仅为2、3、5的正整数。

202. Happy Number

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

12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1

分析：happy number的定义为：将数字的各个位对应的数字平方相加，将所得的数据继续上面的过程，如果出现得到 1 ，那么定义初始数据为“happy number”，若出现不为1的循环，则不是。

由定义即可得到happy number算法，用HashSet来监测循环的产生，如果出现则立即返回false。

public class Solution {
    public boolean isHappy(int n) {
        // 定义法
        int m = n;
        
        Set<Integer> set = new HashSet<Integer>();
        while(m != 1) {
            if(set.contains(m)) return false;// 出现循环就证明不是
            int sum = 0;
            set.add(m);
            
            while(m != 0) { // 每次计算各位<span style="color: rgb(51, 51, 51); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 24px; line-height: 26.3999996185303px;">数字平方相加</span>
                
                sum += Math.pow(m % 10, 2);
                m = m /10;
            }
            m = sum;
        }
        return true;
    }
}

263. Ugly Number

Write a program to check whether a given number is an ugly number.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 6, 8 are ugly while 14 is not ugly since it includes another prime factor 7.

Note that 1 is typically treated as an ugly number.

分析：ugly number 只包含因数 2，3，5. 另外 1 也为 ugly number。

public class Solution {
    public boolean isUgly(int num) {
        if(num == 0) return false;
        while(num % 2 == 0 || num % 3 == 0 || num % 5 == 0){ 
            
            if(num % 2 == 0){
                num = num / 2;
            }else if(num % 3 == 0){
                num = num / 3;
            }else{
                num = num / 5;
            }
        }
        if(num == 1) return true;
        return false;
    }
}