(LeetCode 202) Happy Number-优快云博客

本文链接：https://blog.youkuaiyun.com/dreamgchuan/article/details/51174101

给定一个正整数，通过不断将数字的每位平方和替换原数，若最终结果为1，则该数为Happy Number。若过程中出现重复数字，表明会无限循环，该数不是Happy Number。解决方案利用unordered_map来检测重复项。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Q:
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

题意大概是，给你一个整数，将它替换为他各个位的平方和，如此重复，如果她能够最后变成1.那么他就是Happy Number。否则他就会无限循环下去。

solution:
判断它是Happy Number相信大家都会，可是怎么判定它不是Happy Number呢？
我想了半天，后来发现题目里面已经给了提示，因为他最终会无限循环，就是说最后会有重复的数出现。这里我也学会使用unordered_map了。

class Solution {
public:
    bool isHappy(int n) {
        unordered_map<int,bool>check;
        check[n]=true;
        while(n!=1){
            n = decomposite(n);
            if(check[n]==true)return false;
            check[n]=true;
        }
        return true;
    }
    int decomposite(int n){
        int res=0;
        while(n){
            int m=n%10;
            n/=10;
            res += m*m;
        }
        return res;
    }
};