202. 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

• 12 + 92 = 82
• 82 + 22 = 68
• 62 + 82 = 100
• 12 + 02 + 02 = 1

思路：

本题有两种思路：第一种寻找特殊值，发现无论怎样的数字，要想最后结果为1，除了10的倍数外，其余数字必须求和为70才能保证为欢乐树，否则false;第二种思路就是找个数字计算，找规律，发现经历一定不走计算后，和要么为1要么就经历死循环。

代码（特殊值）：

class Solution {public:    bool isHappy(int n) {        vector<int> a = ret(n);        int sum = res(a);        cout<<sum<<endl;        if(sum==1)            return true;        else if(sum%10==0&&ret(sum).size()==2)        {            if(sum==10||sum==70)                return true;            else                return false;        }        else            return isHappy(sum);          }    vector<int> ret(int n){        vector<int> num;        while(n)        {            num.push_back(n%10);            n = n/10;        }        return num;    }    int res(vector<int> num){        int sum = 0;        for(int i = 0;i<num.size();i++)            sum+=num[i]*num[i];               return sum;    }    };

代码（找规律）：

class Solution {public:int digitSquareSum(int n) {    int sum = 0, tmp;    while (n) {        tmp = n % 10;        sum += tmp * tmp;        n /= 10;    }    return sum;}bool isHappy(int n) {    int slow, fast;    slow = fast = n;    do {        slow = digitSquareSum(slow);        fast = digitSquareSum(fast);        fast = digitSquareSum(fast);    } while(slow != fast);    if (slow == 1) return 1;    else return 0;}};