leetcode--Happy Number_leetcode happynum-优快云博客

本文深入探讨了快乐数字的概念和算法实现，通过多种方法验证一个正整数是否为快乐数，包括迭代求和、哈希集合和特定个位数判断等。详细介绍了不同算法版本及其背后的逻辑。

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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

解题思路：

#include <iostream>
#include <algorithm>
#include <math.h>
using namespace std;



int sum(int n)
{
    int sum=0;
    while(n)
    {
        sum+=pow(n%10,2);
        n /= 10;
    }
    return sum;
}


bool isHappy(int n)
{
    int m=6;
    while(m--)
    {
        n=sum(n);
        if(n==1)
            return true;
    }
    return false;
}



int main()
{
    int n=3;
    if(isHappy(n)) cout<<"true"<<endl;
    else cout<<"false"<<endl;
    return 0;
}

版本二，如果在计算过程中出现了原来出现过的和，那么就一定不能是happynum.基于这个思路使用的是hashset

 public boolean isHappy(int n) {
        Set<Integer> res = new HashSet<>();
        int sum = n;
        while (sum != 1) {
            n = sum;
            sum = 0;
            while (n != 0) {
                sum += (n % 10) * (n % 10);
                n /= 10;
            }
            if (res.contains(sum)) {
                return false;
            }
            res.add(sum);
        }
        return true;
    }

版本三，在整个求和过程中，最后的个位数结果只能是1或者7才能是happynum，其余的都不是，基于这个思路，使用递归的思路进行就是那即可。

public class Solution {
    public boolean isHappy(int n) {
      if(n==1||n==7) return true;
      else if(n<10) return false;
       int result=0;
       while(n > 0){
            int i = n % 10;
            n = n/10;
            result += Math.pow(i,2);
        }
        return isHappy(result);
    }
}