# Java solution, 2 ms, without using Set, beat 84%

• I think the important part is how to end a potential infinite loop. Some suggest using Set, it is a straight forward solution, but I found it will increase the time cost. Therefore, I use another condition to detect an infinite loop, which is a bit tricky though.

``````public class Solution {
public boolean isHappy(int n) {

if ( n < 5) {
if (1 == n)
return true;
else
return false;
}

int sum = 0;

while ( n >= 10) {
sum += (n % 10) * (n % 10);
n = n / 10;
}
sum += n * n;

return  isHappy(sum);
}
}
``````

• @CHYGO same idea here, but I eventually give up this method since I can't even mathematically prove that for every number, the sum is gonna fall below 10, not even to mention 5.

• I had the same performance with below. The exit is based on the fact that converge (if its going to happen) seems to occur within 5 iterations for numbers that don't overflow int.

``````public class Solution {
boolean isHappy(int n) {
return isHappy(n, 0);
}

boolean isHappy(int n, int depth) {
if (n == 1) return true;
if (n == 0 || depth> 5) return false;

int sum = 0;
while(n > 0)
{
int d = (n%10);
sum += d*d;
n/=10;
}
return isHappy(sum, depth+1);
}
}
``````

• @noah.s.moore How can you prove that if the depth is greater than 5,then return false?

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