I'm looking for the fastest way to determine if a long value is a perfect square (i.e. its square root is another integer). I've done it the easy way, by using the built-in Math.sqrt() function, but I'm wondering if there is a way to do it faster by restricting yourself to integer-only domain. Maintaining a lookup table is impratical (since there are about 231.5 integers whose square is less than 263).
Here is the very simple and straightforward way I'm doing it now:
public final static boolean isPerfectSquare(long n)
if (n < 0)
long tst = (long)(Math.sqrt(n) + 0.5);
return tst*tst == n;
Notes: I'm using this function in many Project Euler problems. So no one else will ever have to maintain this code. And this kind of micro-optimization could actually make a difference, since part of the challenge is to do every algorithm in less than a minute, and this function will need to be called millions of times in some problems.