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

基于java解决方案的短php算法返回k元素从n(二项式系数)的所有组合:

$array = array(1,2,3,4,5);

$array_result = NULL;

$array_general = NULL;

function combinations($array, $len, $start_position, $result_array, $result_len, &$general_array)
{
    if($len == 0)
    {
        $general_array[] = $result_array;
        return;
    }

    for ($i = $start_position; $i <= count($array) - $len; $i++)
    {
        $result_array[$result_len - $len] = $array[$i];
        combinations($array, $len-1, $i+1, $result_array, $result_len, $general_array);
    }
} 

combinations($array, 3, 0, $array_result, 3, $array_general);

echo "<pre>";
print_r($array_general);
echo "</pre>";

相同的解决方案，但在javascript:

var newArray = [1, 2, 3, 4, 5];
var arrayResult = [];
var arrayGeneral = [];

function combinations(newArray, len, startPosition, resultArray, resultLen, arrayGeneral) {
    if(len === 0) {
        var tempArray = [];
        resultArray.forEach(value => tempArray.push(value));
        arrayGeneral.push(tempArray);
        return;
    }
    for (var i = startPosition; i <= newArray.length - len; i++) {
        resultArray[resultLen - len] = newArray[i];
        combinations(newArray, len-1, i+1, resultArray, resultLen, arrayGeneral);
    }
} 

combinations(newArray, 3, 0, arrayResult, 3, arrayGeneral);

console.log(arrayGeneral);

简短快速的c#实现

public static IEnumerable<IEnumerable<T>> Combinations<T>(IEnumerable<T> elements, int k)
{
    return Combinations(elements.Count(), k).Select(p => p.Select(q => elements.ElementAt(q)));                
}      

public static List<int[]> Combinations(int setLenght, int subSetLenght) //5, 3
{
    var result = new List<int[]>();

    var lastIndex = subSetLenght - 1;
    var dif = setLenght - subSetLenght;
    var prevSubSet = new int[subSetLenght];
    var lastSubSet = new int[subSetLenght];
    for (int i = 0; i < subSetLenght; i++)
    {
        prevSubSet[i] = i;
        lastSubSet[i] = i + dif;
    }

    while(true)
    {
        //add subSet ad result set
        var n = new int[subSetLenght];
        for (int i = 0; i < subSetLenght; i++)
            n[i] = prevSubSet[i];

        result.Add(n);

        if (prevSubSet[0] >= lastSubSet[0])
            break;

        //start at index 1 because index 0 is checked and breaking in the current loop
        int j = 1;
        for (; j < subSetLenght; j++)
        {
            if (prevSubSet[j] >= lastSubSet[j])
            {
                prevSubSet[j - 1]++;

                for (int p = j; p < subSetLenght; p++)
                    prevSubSet[p] = prevSubSet[p - 1] + 1;

                break;
            }
        }

        if (j > lastIndex)
            prevSubSet[lastIndex]++;
    }

    return result;
}

赶时髦，发布另一个解决方案。这是一个通用的Java实现。输入:(int k)是要选择的元素数量，(List<T> List)是要选择的列表。返回一个组合列表(list < list <T>>)。

public static <T> List<List<T>> getCombinations(int k, List<T> list) {
    List<List<T>> combinations = new ArrayList<List<T>>();
    if (k == 0) {
        combinations.add(new ArrayList<T>());
        return combinations;
    }
    for (int i = 0; i < list.size(); i++) {
        T element = list.get(i);
        List<T> rest = getSublist(list, i+1);
        for (List<T> previous : getCombinations(k-1, rest)) {
            previous.add(element);
            combinations.add(previous);
        }
    }
    return combinations;
}

public static <T> List<T> getSublist(List<T> list, int i) {
    List<T> sublist = new ArrayList<T>();
    for (int j = i; j < list.size(); j++) {
        sublist.add(list.get(j));
    }
    return sublist;
}

假设你的字母数组是这样的:"ABCDEFGH"。你有三个下标(i, j, k)来表示你要用哪个字母来表示当前单词。

A B C D E F G H
^ ^ ^
i j k

首先你改变k，所以下一步看起来像这样:

A B C D E F G H
^ ^   ^
i j   k

如果你到达终点，你继续改变j和k。

A B C D E F G H
^   ^ ^
i   j k

A B C D E F G H
^   ^   ^
i   j   k

一旦j达到G, i也开始变化。

A B C D E F G H
  ^ ^ ^
  i j k

A B C D E F G H
  ^ ^   ^
  i j   k
...

function initializePointers($cnt) {
    $pointers = [];

    for($i=0; $i<$cnt; $i++) {
        $pointers[] = $i;
    }

    return $pointers;     
}

function incrementPointers(&$pointers, &$arrLength) {
    for($i=0; $i<count($pointers); $i++) {
        $currentPointerIndex = count($pointers) - $i - 1;
        $currentPointer = $pointers[$currentPointerIndex];

        if($currentPointer < $arrLength - $i - 1) {
           ++$pointers[$currentPointerIndex];

           for($j=1; ($currentPointerIndex+$j)<count($pointers); $j++) {
                $pointers[$currentPointerIndex+$j] = $pointers[$currentPointerIndex]+$j;
           }

           return true;
        }
    }

    return false;
}

function getDataByPointers(&$arr, &$pointers) {
    $data = [];

    for($i=0; $i<count($pointers); $i++) {
        $data[] = $arr[$pointers[$i]];
    }

    return $data;
}

function getCombinations($arr, $cnt)
{
    $len = count($arr);
    $result = [];
    $pointers = initializePointers($cnt);

    do {
        $result[] = getDataByPointers($arr, $pointers);
    } while(incrementPointers($pointers, count($arr)));

    return $result;
}

$result = getCombinations([0, 1, 2, 3, 4, 5], 3);
print_r($result);

基于https://stackoverflow.com/a/127898/2628125，但更抽象，适用于任何大小的指针。

Haskell中的简单递归算法

import Data.List

combinations 0 lst = [[]]
combinations n lst = do
    (x:xs) <- tails lst
    rest   <- combinations (n-1) xs
    return $ x : rest

我们首先定义特殊情况，即选择零元素。它产生一个单一的结果，这是一个空列表(即一个包含空列表的列表)。

对于n> 0, x遍历列表中的每一个元素xs是x之后的每一个元素。

Rest通过对组合的递归调用从xs中选取n - 1个元素。该函数的最终结果是一个列表，其中每个元素都是x: rest(即对于x和rest的每个不同值，x为头部，rest为尾部的列表)。

> combinations 3 "abcde"
["abc","abd","abe","acd","ace","ade","bcd","bce","bde","cde"]

当然，由于Haskell是懒惰的，列表是根据需要逐渐生成的，因此您可以部分计算指数级的大组合。

> let c = combinations 8 "abcdefghijklmnopqrstuvwxyz"
> take 10 c
["abcdefgh","abcdefgi","abcdefgj","abcdefgk","abcdefgl","abcdefgm","abcdefgn",
 "abcdefgo","abcdefgp","abcdefgq"]