下面是c++中的迭代算法，它不使用STL，也不使用递归，也不使用条件嵌套循环。这样更快，它不执行任何元素交换，也不会给堆栈带来递归负担，还可以通过分别用mallloc()、free()和printf()替换new、delete和std::cout轻松地移植到ANSI C。

如果你想用不同或更长的字母显示元素，那么改变*字母参数以指向不同于"abcdefg"的字符串。

void OutputArrayChar(unsigned int* ka, size_t n, const char *alphabet) {
    for (int i = 0; i < n; i++)
        std::cout << alphabet[ka[i]] << ",";
    std::cout << endl;
}
    

void GenCombinations(const unsigned int N, const unsigned int K, const char *alphabet) {
    unsigned int *ka = new unsigned int [K];  //dynamically allocate an array of UINTs
    unsigned int ki = K-1;                    //Point ki to the last elemet of the array
    ka[ki] = N-1;                             //Prime the last elemet of the array.
    
    while (true) {
        unsigned int tmp = ka[ki];  //Optimization to prevent reading ka[ki] repeatedly

        while (ki)                  //Fill to the left with consecutive descending values (blue squares)
            ka[--ki] = --tmp;
        OutputArrayChar(ka, K, alphabet);
    
        while (--ka[ki] == ki) {    //Decrement and check if the resulting value equals the index (bright green squares)
            OutputArrayChar(ka, K, alphabet);
            if (++ki == K) {      //Exit condition (all of the values in the array are flush to the left)
                delete[] ka;
                return;
            }                   
        }
    }
}
    

int main(int argc, char *argv[])
{
    GenCombinations(7, 4, "abcdefg");
    return 0;
}

重要提示:字母参数*必须指向至少N个字符的字符串。你也可以传递一个在其他地方定义的字符串地址。

组合:从“7选4”中选择。

下面是一个简单易懂的递归c++解决方案:

#include<vector>
using namespace std;

template<typename T>
void ksubsets(const vector<T>& arr, unsigned left, unsigned idx,
    vector<T>& lst, vector<vector<T>>& res)
{
    if (left < 1) {
        res.push_back(lst);
        return;
    }
    for (unsigned i = idx; i < arr.size(); i++) {
        lst.push_back(arr[i]);
        ksubsets(arr, left - 1, i + 1, lst, res);
        lst.pop_back();
    }
}

int main()
{
    vector<int> arr = { 1, 2, 3, 4, 5 };
    unsigned left = 3;
    vector<int> lst;
    vector<vector<int>> res;
    ksubsets<int>(arr, left, 0, lst, res);
    // now res has all the combinations
}

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"]

static IEnumerable<string> Combinations(List<string> characters, int length)
{
    for (int i = 0; i < characters.Count; i++)
    {
        // only want 1 character, just return this one
        if (length == 1)
            yield return characters[i];

        // want more than one character, return this one plus all combinations one shorter
        // only use characters after the current one for the rest of the combinations
        else
            foreach (string next in Combinations(characters.GetRange(i + 1, characters.Count - (i + 1)), length - 1))
                yield return characters[i] + next;
    }
}

说了这么多，做了这么多，这就是奥卡姆的代码。 算法是显而易见的代码..

let combi n lst =
    let rec comb l c =
        if( List.length c = n) then [c] else
        match l with
        [] -> []
        | (h::t) -> (combi t (h::c))@(combi t c)
    in
        combi lst []
;;

还有另一个递归解决方案(你应该能够使用字母而不是数字)使用堆栈，虽然比大多数更短:

stack = [] 
def choose(n,x):
   r(0,0,n+1,x)

def r(p, c, n,x):
   if x-c == 0:
      print stack
      return

   for i in range(p, n-(x-1)+c):
      stack.append(i)
      r(i+1,c+1,n,x)
      stack.pop()

4选3或者我想要从0到4的所有3种数字组合

choose(4,3) 

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