LeetCode 202. Happy Number

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本文介绍了一种用于判断一个数是否为快乐数的算法。快乐数是指通过将该数的每一位平方和重复替换该数，最终能到达1的正整数。文章提供了两种解题思路：一是使用HashSet存储中间结果避免循环；二是利用规律判断结果是否为1或4来确定是否为快乐数。

题目描述
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:

Input: 19
Output: true
Explanation:
1² + 9² = 82
8² + 2² = 68
6² + 8² = 100
1² + 0² + 0² = 1
解题思路
思路一：按步骤一步步往下计算，用hashset存储中间计算结果，如果最后结果为1，则为happy number；如果循环，则不是happy number。
思路二：找出规律，如果结果为1，则是happy number；如果为4，则不是happy number。
代码

import java.util.HashSet;
public class HappyNumber {
	    public boolean isHappy(int n) {
        Set<Integer> set = new HashSet<>();
		int squareSum = 0, remain = 0;
		while(n != 1) {
			if(set.contains(n))
				return false;
			else 
				set.add(n);
			n = sum(n);
		}
		return true;
    }
    public int sum(int n){
        int Sum = 0, remain = 0;
        while(n != 0){
			remain = n % 10;
			Sum += remain * remain;
			n = n / 10;
		}
        return Sum;
    }
}

class Solution {
    public boolean isHappy(int n) {
        while(true) {
            if(n==4) return false;
            if(n==1) return true;
            n = nextHappy(n);
        }
    }
    private int nextHappy(int n) {
        int sum = 0;
        int lastDigit;
        while(n!=0) {
            lastDigit = n % 10;
            n /= 10;
            sum += lastDigit * lastDigit;
        }
        return sum;
    }
}