另一个具有组合索引惰性生成的c#版本。这个版本维护了一个索引数组来定义所有值列表和当前组合值之间的映射，即在整个运行时不断使用O(k)额外的空间。该代码在O(k)时间内生成单个组合，包括第一个组合。

public static IEnumerable<T[]> Combinations<T>(this T[] values, int k)
{
    if (k < 0 || values.Length < k)
        yield break; // invalid parameters, no combinations possible

    // generate the initial combination indices
    var combIndices = new int[k];
    for (var i = 0; i < k; i++)
    {
        combIndices[i] = i;
    }

    while (true)
    {
        // return next combination
        var combination = new T[k];
        for (var i = 0; i < k; i++)
        {
            combination[i] = values[combIndices[i]];
        }
        yield return combination;

        // find first index to update
        var indexToUpdate = k - 1;
        while (indexToUpdate >= 0 && combIndices[indexToUpdate] >= values.Length - k + indexToUpdate)
        {
            indexToUpdate--;
        }

        if (indexToUpdate < 0)
            yield break; // done

        // update combination indices
        for (var combIndex = combIndices[indexToUpdate] + 1; indexToUpdate < k; indexToUpdate++, combIndex++)
        {
            combIndices[indexToUpdate] = combIndex;
        }
    }
}

测试代码:

foreach (var combination in new[] {'a', 'b', 'c', 'd', 'e'}.Combinations(3))
{
    System.Console.WriteLine(String.Join(" ", combination));
}

输出:

a b c
a b d
a b e
a c d
a c e
a d e
b c d
b c e
b d e
c d e

这是我用c++写的命题

我尽可能少地限制迭代器类型，所以这个解决方案假设只有前向迭代器，它可以是const_iterator。这应该适用于任何标准容器。在参数没有意义的情况下，它抛出std:: invalid_argument

#include <vector>
#include <stdexcept>

template <typename Fci> // Fci - forward const iterator
std::vector<std::vector<Fci> >
enumerate_combinations(Fci begin, Fci end, unsigned int combination_size)
{
    if(begin == end && combination_size > 0u)
        throw std::invalid_argument("empty set and positive combination size!");
    std::vector<std::vector<Fci> > result; // empty set of combinations
    if(combination_size == 0u) return result; // there is exactly one combination of
                                              // size 0 - emty set
    std::vector<Fci> current_combination;
    current_combination.reserve(combination_size + 1u); // I reserve one aditional slot
                                                        // in my vector to store
                                                        // the end sentinel there.
                                                        // The code is cleaner thanks to that
    for(unsigned int i = 0u; i < combination_size && begin != end; ++i, ++begin)
    {
        current_combination.push_back(begin); // Construction of the first combination
    }
    // Since I assume the itarators support only incrementing, I have to iterate over
    // the set to get its size, which is expensive. Here I had to itrate anyway to  
    // produce the first cobination, so I use the loop to also check the size.
    if(current_combination.size() < combination_size)
        throw std::invalid_argument("combination size > set size!");
    result.push_back(current_combination); // Store the first combination in the results set
    current_combination.push_back(end); // Here I add mentioned earlier sentinel to
                                        // simplyfy rest of the code. If I did it 
                                        // earlier, previous statement would get ugly.
    while(true)
    {
        unsigned int i = combination_size;
        Fci tmp;                            // Thanks to the sentinel I can find first
        do                                  // iterator to change, simply by scaning
        {                                   // from right to left and looking for the
            tmp = current_combination[--i]; // first "bubble". The fact, that it's 
            ++tmp;                          // a forward iterator makes it ugly but I
        }                                   // can't help it.
        while(i > 0u && tmp == current_combination[i + 1u]);

        // Here is probably my most obfuscated expression.
        // Loop above looks for a "bubble". If there is no "bubble", that means, that
        // current_combination is the last combination, Expression in the if statement
        // below evaluates to true and the function exits returning result.
        // If the "bubble" is found however, the ststement below has a sideeffect of 
        // incrementing the first iterator to the left of the "bubble".
        if(++current_combination[i] == current_combination[i + 1u])
            return result;
        // Rest of the code sets posiotons of the rest of the iterstors
        // (if there are any), that are to the right of the incremented one,
        // to form next combination

        while(++i < combination_size)
        {
            current_combination[i] = current_combination[i - 1u];
            ++current_combination[i];
        }
        // Below is the ugly side of using the sentinel. Well it had to haave some 
        // disadvantage. Try without it.
        result.push_back(std::vector<Fci>(current_combination.begin(),
                                          current_combination.end() - 1));
    }
}

我有一个用于project euler的排列算法，用python编写:

def missing(miss,src):
    "Returns the list of items in src not present in miss"
    return [i for i in src if i not in miss]


def permutation_gen(n,l):
    "Generates all the permutations of n items of the l list"
    for i in l:
        if n<=1: yield [i]
        r = [i]
        for j in permutation_gen(n-1,missing([i],l)):  yield r+j

If

n<len(l) 

你应该有所有你需要的组合，没有重复，你需要吗?

它是一个生成器，所以你可以这样使用它:

for comb in permutation_gen(3,list("ABCDEFGH")):
    print comb 

一个简洁的Javascript解决方案:

Array.prototype.combine=function combine(k){    
    var toCombine=this;
    var last;
    function combi(n,comb){             
        var combs=[];
        for ( var x=0,y=comb.length;x<y;x++){
            for ( var l=0,m=toCombine.length;l<m;l++){      
                combs.push(comb[x]+toCombine[l]);           
            }
        }
        if (n<k-1){
            n++;
            combi(n,combs);
        } else{last=combs;}
    }
    combi(1,toCombine);
    return last;
}
// Example:
// var toCombine=['a','b','c'];
// var results=toCombine.combine(4);

JavaScript，基于生成器，递归方法:

function *nCk(n,k){ for(var i=n-1;i>=k-1;--i) if(k===1) yield [i]; else for(var temp of nCk(i,k-1)){ temp.unshift(i); yield temp; } } function test(){ try{ var n=parseInt(ninp.value); var k=parseInt(kinp.value); log.innerText=""; var stop=Date.now()+1000; if(k>=1) for(var res of nCk(n,k)) if(Date.now()<stop) log.innerText+=JSON.stringify(res)+" "; else{ log.innerText+="1 second passed, stopping here."; break; } }catch(ex){} } n:<input id="ninp" oninput="test()"> &gt;= k:<input id="kinp" oninput="test()"> &gt;= 1 <div id="log"></div>

通过这种方式(减少i和unshift())，它以递减的顺序生成组合和组合内的元素，有点赏心悦目。 测试在1秒后停止，因此输入奇怪的数字是相对安全的。

不需要进行集合操作。这个问题几乎和循环K个嵌套循环一样，但你必须小心索引和边界(忽略Java和OOP的东西):

 public class CombinationsGen {
    private final int n;
    private final int k;
    private int[] buf;

    public CombinationsGen(int n, int k) {
        this.n = n;
        this.k = k;
    }

    public void combine(Consumer<int[]> consumer) {
        buf = new int[k];
        rec(0, 0, consumer);
    }

    private void rec(int index, int next, Consumer<int[]> consumer) {
        int max = n - index;

        if (index == k - 1) {
            for (int i = 0; i < max && next < n; i++) {
                buf[index] = next;
                next++;
                consumer.accept(buf);
            }
        } else {
            for (int i = 0; i < max && next + index < n; i++) {
                buf[index] = next;
                next++;
                rec(index + 1, next, consumer);
            }
        }
    }
}

像这样使用:

 CombinationsGen gen = new CombinationsGen(5, 2);

 AtomicInteger total = new AtomicInteger();
 gen.combine(arr -> {
     System.out.println(Arrays.toString(arr));
     total.incrementAndGet();
 });
 System.out.println(total);

获得预期的结果:

[0, 1]
[0, 2]
[0, 3]
[0, 4]
[1, 2]
[1, 3]
[1, 4]
[2, 3]
[2, 4]
[3, 4]
10

最后，将索引映射到您可能拥有的任何数据集。