Quickest way to generate a sequence of consecutive numbers in a random order

Question

Whats the quickest way to generate a list of consecutive numbers in a random order?

i.e. generate a list of numbers from 1 to 100, the list must contain each number once only. The order of the list should be random.

java or c# please.

my pseudocode looks like this, very inefficient.

 var list = new list<int>();
 for (int i = 1; i <= 100; ++i) {
     int x;
     repeat {
         x = random(1, 100);
     until (list.contains(x) == false);
     list.add(x);
 }

The usual solution is to generate an ordered array then scramble it. — Denys Séguret, Feb 24 '14 at 15:51
`List list = Enumerable.Range(1, 100).ToList();` and then shuffle it like [this answer](http://stackoverflow.com/questions/273313/randomize-a-listt-in-c-sharp) — Habib, Feb 24 '14 at 15:51
@Tarec, yes, just posted it because those will be the only numbers possible in the list. — Habib, Feb 24 '14 at 15:52
You're looking for the Fisher-Yates algorithm. http://stackoverflow.com/questions/1150646/card-shuffling-in-c-sharp — Robert Harvey, Feb 24 '14 at 15:53
You just gotta try hard enough and once in a while the order will be 1 to 100. — Marco, Feb 24 '14 at 15:53

score 3 · Accepted Answer · answered Feb 24 '14 at 15:52

3

Yes, it's teribly inefficient and it's not even bounded in time.

The usual solution is

generate an ordered array
shuffle it (I'd recommend the Fisher–Yates shuffle, it's simple, fast, unbiased and you'll find an implementation in any language)

answered Feb 24 '14 at 15:52

Denys Séguret

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Quickest way to generate a sequence of consecutive numbers in a random order

1 Answers1