问题
This was inspired by a question at a job interview: how do you efficiently generate N unique random numbers? Their security and distribution/bias don't matter.
I proposed a naive way of calling rand() N times and eliminating dupes by trial and error, thus getting inefficient and flawed solution. Then I've read this SO question, these algorithms are great for getting quality unique numbers and they are O(N).
But I suspect there are ways to get low-quality unique random numbers for dummy tasks in less than O(N) time complexity. I got some possible ideas:
- Store many precomputed lists each containing N numbers and retrieve one list randomly. Complexity is O(1) for fixed N. Storage space used is O(NR) where R is number of lists.
- Generate N/2 unique random numbers and then divide them by 2 inequal parts (floor/ceil for odd numbers, n+1/n-1 for even). I know this is flawed (duplicates can pop up) and O(N/2) is still O(N). This is more of a food for thought.
- Generate one big random number and then squeeze more variants from it by some fixed manipulations like bitwise operations, factorization, recursion, MapReduce or something else.
- Use a quasi-random sequence somehow (not a math guy, just googled this term).
Your ideas?
回答1:
Presumably this routine has some kind of output (i.e. the results are written to an array of some kind). Populating an array (or some other data-structure) of size N is at least an O(N) operation, so you can't do better than O(N).
回答2:
You can consequently generate a random number, and if the result array contains it, just add to it the maximum number of already generated numbers.
Detecting if a number already generated is O(1) (using a hash set). So it's O(n) and with only N random() calls.
Of course, this is an assumption that we do not overflow the upper limit (i.e. BigInteger).
来源：https://stackoverflow.com/questions/10064550/generate-n-quasi-random-numbers-in-less-than-on