C ++ 11 int x，int y之间的随机数，不包括列表<int>

Question

I need a way to generate random integers between x,y, but once a random z is generated, I need the next iteration of x,y to exclude z (or better yet, a list of ints to exclude). The return value has to honor the "std::uniform_int_distribution<>" condition. There is an obvious way to do this, but I was hoping for a fast version.

int GenerateRandomBetweenExcluding(int min, int max, int exclude) // even better is a list<int> to exclude
{
    std::random_device rd;
    std::mt19937 gen(rd());
    std::uniform_int_distribution<> dis(min, max); // how to exclude "exclude" or list<int>?

    int randNum = dis(gen);

    return randNum;
}

Answer 1

Use a list with all numbers between min and max. Choose randomly an element and delete it.

Answer 2

#include <algorithm>
#include <vector>
#include <random>

class NoRepeat
{
public:
    void Init(int,int);
    int GetOne();
    bool HasMore();
private:
    std::vector<int> list;
};

void NoRepeat::Init(int min, int max) // min and max are inclusive
{
    list.clear();
    list.reserve(max-min+1);
    for( int i=min; i<=max; ++i )
        list.push_back(i);
    std::random_shuffle( list.begin(), list.end() );
    // might want to supply own random_func, default uses rand()
}

bool NoRepeat::HasMore()
{
    return !list.empty();
}

int NoRepeat::GetOne()
{
    int ret = list.back();
    list.pop_back();
    return ret;
}

Answer 3

There's a nice way to do this if your exclusion list is sorted in ascending order. You can achieve this sorting by using a set<int> instead of a list<int> to track exclusions; most set operations take log(n) time and the container is guaranteed to be sorted at all times.

The algorithm is as follows:

set<int> exclude;
/* ... configure generator and exclusion set ... */
uniform_int_distribution<> dis(min, max-exclude.size());
int val = dis(gen);

/* skip over the exclusions */
for(int i : exclude) {
    if(val >= i) val++;
    else break;
}
return val;