LeetCode 202:Happy Number

最新推荐文章于 2021-09-12 12:51:45 发布

zhihua_bupt

最新推荐文章于 2021-09-12 12:51:45 发布

阅读量1.3k

点赞数 1

CC 4.0 BY-SA版权

分类专栏： LeetCode C++ DataStructure&Algorithms LeetCode算法分析

本文链接：https://blog.youkuaiyun.com/geekmanong/article/details/50279915

C++ 同时被 3 个专栏收录

133 篇文章

订阅专栏

DataStructure&Algorithms

131 篇文章

订阅专栏

LeetCode算法分析

128 篇文章

订阅专栏

本文介绍了一种用于判断任意正整数是否为快乐数的算法。快乐数是指通过反复替换为各位数字平方和的过程最终能够得到1的正整数。文章详细展示了如何使用C++实现这一算法，并附带了一个具体的例子来说明判断过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

//利用set记录每次得到的平方和，当出现非1的重复平方和时，返回False,否则，返回True
class Solution {
public:
	bool isHappy(int n) {
		set<int> set1;
		if (n <= 0) return false;
		if (n == 1) return true;

		int sum = 0;
		int p = 0;
		while (n!=1 ){
			while (n){
				 p = n % 10;
				 n = n / 10;
				sum = sum + p*p;
			}
			if (set1.find(sum)!=set1.end()) 
				return false;
			n = sum;
			sum = 0;
			set1.insert(n);
		}
		return n == 1;
	}
};