《计算机编程艺术，卷4A:组合算法，第1部分》第7.2.1.3节中算法L(字典组合)的C代码:

#include <stdio.h>
#include <stdlib.h>

void visit(int* c, int t) 
{
  // for (int j = 1; j <= t; j++)
  for (int j = t; j > 0; j--)
    printf("%d ", c[j]);
  printf("\n");
}

int* initialize(int n, int t) 
{
  // c[0] not used
  int *c = (int*) malloc((t + 3) * sizeof(int));

  for (int j = 1; j <= t; j++)
    c[j] = j - 1;
  c[t+1] = n;
  c[t+2] = 0;
  return c;
}

void comb(int n, int t) 
{
  int *c = initialize(n, t);
  int j;

  for (;;) {
    visit(c, t);
    j = 1;
    while (c[j]+1 == c[j+1]) {
      c[j] = j - 1;
      ++j;
    }
    if (j > t) 
      return;
    ++c[j];
  }
  free(c);
}

int main(int argc, char *argv[])
{
  comb(5, 3);
  return 0;
}

下面是我的Scala解决方案:

def combinations[A](s: List[A], k: Int): List[List[A]] = 
  if (k > s.length) Nil
  else if (k == 1) s.map(List(_))
  else combinations(s.tail, k - 1).map(s.head :: _) ::: combinations(s.tail, k)

下面是Clojure版本，它使用了我在OCaml实现答案中描述的相同算法:

(defn select
  ([items]
     (select items 0 (inc (count items))))
  ([items n1 n2]
     (reduce concat
             (map #(select % items)
                  (range n1 (inc n2)))))
  ([n items]
     (let [
           lmul (fn [a list-of-lists-of-bs]
                     (map #(cons a %) list-of-lists-of-bs))
           ]
       (if (= n (count items))
         (list items)
         (if (empty? items)
           items
           (concat
            (select n (rest items))
            (lmul (first items) (select (dec n) (rest items))))))))) 

它提供了三种调用方法:

(a)按问题要求，选出n项:

  user=> (count (select 3 "abcdefgh"))
  56

(b) n1至n2个选定项目:

user=> (select '(1 2 3 4) 2 3)
((3 4) (2 4) (2 3) (1 4) (1 3) (1 2) (2 3 4) (1 3 4) (1 2 4) (1 2 3))

(c)在0至所选项目的集合大小之间:

user=> (select '(1 2 3))
(() (3) (2) (1) (2 3) (1 3) (1 2) (1 2 3))

这是我对javascript的贡献(没有递归)

set = ["q0", "q1", "q2", "q3"]
collector = []


function comb(num) {
  results = []
  one_comb = []
  for (i = set.length - 1; i >= 0; --i) {
    tmp = Math.pow(2, i)
    quotient = parseInt(num / tmp)
    results.push(quotient)
    num = num % tmp
  }
  k = 0
  for (i = 0; i < results.length; ++i)
    if (results[i]) {
      ++k
      one_comb.push(set[i])
    }
  if (collector[k] == undefined)
    collector[k] = []
  collector[k].push(one_comb)
}


sum = 0
for (i = 0; i < set.length; ++i)
  sum += Math.pow(2, i)
 for (ii = sum; ii > 0; --ii)
  comb(ii)
 cnt = 0
for (i = 1; i < collector.length; ++i) {
  n = 0
  for (j = 0; j < collector[i].length; ++j)
    document.write(++cnt, " - " + (++n) + " - ", collector[i][j], "<br>")
  document.write("<hr>")
}   

像Andrea Ambu一样用Python写的，但不是硬编码来选择三个。

def combinations(list, k):
    """Choose combinations of list, choosing k elements(no repeats)"""
    if len(list) < k:
        return []
    else:
        seq = [i for i in range(k)]
        while seq:
            print [list[index] for index in seq]
            seq = get_next_combination(len(list), k, seq)

def get_next_combination(num_elements, k, seq):
        index_to_move = find_index_to_move(num_elements, seq)
        if index_to_move == None:
            return None
        else:
            seq[index_to_move] += 1

            #for every element past this sequence, move it down
            for i, elem in enumerate(seq[(index_to_move+1):]):
                seq[i + 1 + index_to_move] = seq[index_to_move] + i + 1

            return seq

def find_index_to_move(num_elements, seq):
        """Tells which index should be moved"""
        for rev_index, elem in enumerate(reversed(seq)):
            if elem < (num_elements - rev_index - 1):
                return len(seq) - rev_index - 1
        return None   

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。