std :: mt19937 mersenne twister分布与非重复值

Question

I'd like to use the std::mt19937 random number generator to produce a list of numbers between 0 and 255. "Once a number has been chosen, it should not appear again in the set." - it's this bit I don't know how to do. The mathematical term for this escapes me(!)

std::mt19937                        twister;
std::uniform_int_distribution<int>  distribution;

twister.seed(91210);
distribution = std::uniform_int_distribution<int>(0,255);

std::vector vNumbers;
vNumbers.resize(256);

for( int n = 0; n < 256; ++ n )
    vNumbers[n] = distribution(twister);

Answer 1

There's algorithms for that:

// fill a vector ith [0..255]:
std::vector<int> vNumbers(256);
std::iota(vNumbers.begin(), vNumbers.end(), 0);

// shuffle it
std::random_shuffle(vNumbers.begin(), vNumbers.end());

// done

With C++11 you can pass in your own generator for the RNG: (see also comments)

std::shuffle(vNumbers.begin(), vNumbers.end(), twister);

Or you could roll your own (google Fisher-Yates, or see Knuth)

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Of course the iota can be replaced by te following

for (int i=0; i<256; ++i) vNumbers[i] = i;

Answer 2

You could use std::set instead of std::vector , as it will ensure you have no duplicates. Just loop until the size of the set is the number of value you want.

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Since it seems you want one of each number in just a random order, you don't really need to generate random numbers, just generate a sequence from 0 to 255, and then randomly shuffle them.

This can easily be done with some standard algorithms , like std::iota and std::random_shuffle :

std::vector<int> values{256};
std::iota(values.begin(), values.end(), 0);
std::random_shuffle(values.begin(), values.end());