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class CollatzCalculator {
int computeStepCount(int start) {
if (start < 1) {
throw new IllegalArgumentException("Only positive integers are allowed");
}
int steps = 0;
while (start > 1) {
if ((start & 1) == 1) {
start = start * 3 + 1;
} else {
start >>= 1;
}
steps++;
}
return steps;
}
}
This approach defines a steps variable initialized to 0 that will be incremented for each iteration of the while loop.
The while loop iterates as long as the input number is greater than 1.
If the number is already 1 before control flow gets to the while loop, then the while loop won't run.
Bitwise operators are used to check if the number is odd and to divide it in half if it is even.
The bitwise AND operator (&) compares the number with 1 to see if it is odd.
Another way to go about checking if the number is even or odd is to see if it is evenly divided by 2
by using the modulo (also know as the remainder) operator.
So, to see if it is even we could use if (start % 2 == 0).
That might be slightly less performant but perhaps more readable for intent, especially for those who don't "speak binary".
If the number is even, then the right shift equals operator (>>=) shifts all of the bits once to the right, which is the equivalent
of dividing the number by 2.
Another way would be to use the division equals operator start \= 2 which might be slighly less performant but more readable
for intent, especially for those who don't "speak binary".
After the condition for the while loop is satisifed, it ends and the value of steps is returned.
