Leetcode 快乐数Happy Number

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

一开始的想法就是直接按定义来：

int getsum(int x){
int sum=0;
while(x){
sum+=(x%10)*(x%10);
x=x/10;}
return sum;
}
bool isHappy(int n) {
if(n<=0)return false;
bool f=1;
long k=getsum(n);
while(k!=1){
f=0;
k=getsum(k);}
if(f==0)return true;
else return false;
}

最后发现运行时间够长，Time Limit Exceeded
最终查阅有关快来数定义，发现当在迭代循环判断是否为1的过程中时，若进入了这种无限循环，最终都会出现4，故一旦遇见4即可知道进入了无限循环，判断是非快乐数字，减少了时间复杂度：

int getsum(int x){
int sum=0;
while(x){
sum+=(x%10)*(x%10);
x=x/10;}
return sum;
}
bool isHappy(int n) {
if(n<=0)return false;

while(n!=1){
n=getsum(n);
if(n==4)return false;}
return true;
}