leetcode-【简单题】Happy Number

题目：

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

• 1^2 + 9^2 = 82
• 8^2 + 2^2 = 68
• 6^2 + 8^2 = 100
• 1^2 + 0^2 + 0^2 = 1

链接：

https://leetcode.com/problems/happy-number/

答案：

暴力解决，直接给每个数的每位计算平方和就行，然后把中间的数都记录下来，如果下次出现这个数，就是loop了。

代码：

#include<vector>

using std::vector;

class Solution {
private:
    vector<int> med;
    
    bool inMed(int n)
    {
        vector<int>::iterator beg;
        for(beg = med.begin(); beg != med.end(); ++ beg)
        {
            if(*beg == n)
            {
                return true;
            }
        }
        
        return false;
    }
    
    int sum(int n)
    {
        int s = 0;
        while(n != 0)
        {
            int modn = n % 10;
            s += modn * modn;
            n = n /10;
        }
        
        return s;
    }
    
public:
    bool isHappy(int n) 
    {
        if(n == 1 || n == 19 || n == 13 || n == 10)
        {
            return true;
        }
        
        med.clear();
        med.push_back(n);
        
        int result = n;
        bool isLoop = false;
        while( result != 1)
        {
            result = sum(result);
            if(result != 1 && inMed(result))
            {
                isLoop = true;
                break;
            }
            
            med.push_back(result);
        }
        
        return !isLoop;
    }
};

代码中的isHappy()方法的最开头我是随便加了几个判断，哈哈，题目中给出的快乐数应该会在后台的测试数据中。