LinkedList implements it with a doubly-linked list. LinkedList and ArrayList are two different implementations of the List interface. In LinkedList adding an element takes O(n) time and accessing also takes O(n) time but LinkedList uses more memory than ArrayList. TLDR, in ArrayList accessing an element takes constant time and adding an element takes O(n) time. If you're not sure - just start with ArrayList. Summary ArrayList with ArrayDeque are preferable in many more use-cases than LinkedList. Compute sqrt using something like Hensel's lemma Divide out powers of 4 using binary search In my actual code, I use the following modified loop: This is true even modulo numbers, but beware, modulo some numbers, things can have even more than 2 square roots notably, this includes powers of 2.) Because our actual square root is less than 2^32, at that point we can actually just check if r or t/2-r are real square roots. (Note that if r is a square root of x, then so is -r. At the end, r and t/2-r will be square roots of x modulo t/2. The idea is that at each iteration, you add one bit onto r, the "current" square root of x each square root is accurate modulo a larger and larger power of 2, namely t/2. (Note: untested code if it doesn't work, try t=2 or 8.) The basic structure of Hensel's lemma is the following. (I don't think it's applicable directly, but it works with some modifications.) Before doing that, I divide out all powers of 2 with a binary search:Īt this point, for our number to be a square, it must be 1 mod 8. Finally, try to compute the square root using a method similar to Hensel's lemma.However, I just use a table of size 512 To actually check if the residue is a square, I look up the answer in a precomputed table. At this point, y is between 0 and 511. (In reading the code below, note that my input is int64 x.) (I found looking at the last six didn't help.) I also answer yes for 0. This includes negative numbers and looking at the last 4 bits. Your results may vary, especially because I don't know how the Java factor will play out. I figured out a method that works ~35% faster than your 6bits+Carmack+sqrt code, at least with my CPU (x86) and programming language (C/C++).
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