从n返回k个元素的所有组合的算法
我想写一个函数，它将一个字母数组作为参数，并选择一些字母。
假设您提供了一个包含8个字母的数组，并希望从中选择3个字母。然后你应该得到：
8! / ((8 - 3)! * 3!) = 56
返回的数组（或单词），每个包含3个字母。
77 个答案:
答案 0 :(得分：395)
Art of Computer Programming Volume 4: Fascicle 3有很多这些可能比我描述的更适合你的特殊情况。
格雷码
你会遇到的一个问题当然是记忆而且非常快，你的集合中有20个元素会有问题 - 20 C 3 = 1140如果你想迭代这个集合，最好使用修改后的格雷码算法，这样你就不会把所有这些都保存在内存中。这些产生了前一个的下一个组合，避免重复。其中有许多用于不同的用途。我们是否希望最大化连续组合之间的差异？最小化？等等。
一些描述格雷码的原始论文：
- Some Hamilton Paths and a Minimal Change Algorithm
- Adjacent Interchange Combination Generation Algorithm
- An Efficient Implementation of the Eades, Hickey, Read Adjacent Interchange Combination Generation Algorithm（PDF，代码为Pascal）
- Combination Generators
- Survey of Combinatorial Gray Codes（PostScript）
- An Algorithm for Gray Codes
- 由于组合是无序的，{1,3,2} = {1,2,3} - 我们将它们命名为词典。
- 此方法有一个隐式0来启动第一个差异的设置。
以下是其他一些涉及该主题的论文：
Chase's Twiddle（算法）
Phillip J Chase，“Algorithm 382: Combinations of M out of N Objects”（1970）
字典顺序中的组合索引（带扣算法515）
您还可以通过其索引（按字典顺序）引用组合。根据索引意识到索引应该是从右到左的一些变化，我们可以构造一些应该恢复组合的东西。
所以，我们有一套{1,2,3,4,5,6} ......我们想要三个元素。让我们说{1,2,3}我们可以说元素之间的差异是一个，有序和最小。 {1,2,4}有一个变化，按字典顺序排列第2位。因此，最后一个地方的“变化”数量占字典顺序的一个变化。第二个位置，只有一个变化{1,3,4}有一个变化，但由于它位于第二位（与原始集合中的元素数量成比例），因此会有更多变化。
我所描述的方法是解构，看起来，从设置到索引，我们需要做反向 - 这更加棘手。这就是Buckles解决问题的方法。我写了一些C to compute them，稍作修改 - 我使用集合的索引而不是数字范围来表示集合，所以我们总是从0 ... n开始工作。 注意：
字典顺序中的组合索引（McCaffrey）
有another way：，它的概念更易于掌握和编程，但它没有Buckles的优化。幸运的是，它也不会产生重复的组合：
最大化
的
集
。
例如：27 = C(6,4) + C(5,3) + C(2,2) + C(1,1)。因此，第四十七个词典组合的四个事物是：{1,2,5,6}，这些是你想要看的任何集合的索引。下面的示例（OCaml）需要choose函数，留给读者：
(* this will find the [x] combination of a [set] list when taking [k] elements *)
let combination_maccaffery set k x =
(* maximize function -- maximize a that is aCb *)
(* return largest c where c < i and choose(c,i) <= z *)
let rec maximize a b x =
if (choose a b ) <= x then a else maximize (a-1) b x
in
let rec iterate n x i = match i with
| 0 -> []
| i ->
let max = maximize n i x in
max :: iterate n (x - (choose max i)) (i-1)
in
if x < 0 then failwith "errors" else
let idxs = iterate (List.length set) x k in
List.map (List.nth set) (List.sort (-) idxs)
一个小而简单的组合迭代器
提供以下两种算法用于教学目的。它们实现了迭代器和（更一般的）文件夹整体组合。
它们尽可能快，具有复杂度O（ n C k ）。内存消耗受k的约束。
我们将从迭代器开始，它将为每个组合调用用户提供的函数
let iter_combs n k f =
let rec iter v s j =
if j = k then f v
else for i = s to n - 1 do iter (i::v) (i+1) (j+1) done in
iter [] 0 0
更通用的版本将从初始状态开始调用用户提供的函数以及状态变量。因为我们需要在不同状态之间传递状态，所以我们不会使用for循环，而是使用递归，
let fold_combs n k f x =
let rec loop i s c x =
if i < n then
loop (i+1) s c @@
let c = i::c and s = s + 1 and i = i + 1 in
if s < k then loop i s c x else f c x
else x in
loop 0 0 [] x
答案 1 :(得分：186)
在C＃中：
public static IEnumerable<IEnumerable<T>> Combinations<T>(this IEnumerable<T> elements, int k)
{
return k == 0 ? new[] { new T[0] } :
elements.SelectMany((e, i) =>
elements.Skip(i + 1).Combinations(k - 1).Select(c => (new[] {e}).Concat(c)));
}
用法：
var result = Combinations(new[] { 1, 2, 3, 4, 5 }, 3);
结果：
123
124
125
134
135
145
234
235
245
345
答案 2 :(得分：74)
简短的java解决方案：
import java.util.Arrays;
public class Combination {
public static void main(String[] args){
String[] arr = {"A","B","C","D","E","F"};
combinations2(arr, 3, 0, new String[3]);
}
static void combinations2(String[] arr, int len, int startPosition, String[] result){
if (len == 0){
System.out.println(Arrays.toString(result));
return;
}
for (int i = startPosition; i <= arr.length-len; i++){
result[result.length - len] = arr[i];
combinations2(arr, len-1, i+1, result);
}
}
}
结果将是
[A, B, C]
[A, B, D]
[A, B, E]
[A, B, F]
[A, C, D]
[A, C, E]
[A, C, F]
[A, D, E]
[A, D, F]
[A, E, F]
[B, C, D]
[B, C, E]
[B, C, F]
[B, D, E]
[B, D, F]
[B, E, F]
[C, D, E]
[C, D, F]
[C, E, F]
[D, E, F]
答案 3 :(得分：73)
我可以提出我的递归Python解决方案来解决这个问题吗？
def choose_iter(elements, length):
for i in xrange(len(elements)):
if length == 1:
yield (elements[i],)
else:
for next in choose_iter(elements[i+1:len(elements)], length-1):
yield (elements[i],) + next
def choose(l, k):
return list(choose_iter(l, k))
使用示例：
>>> len(list(choose_iter("abcdefgh",3)))
56
我喜欢它的简单。
答案 4 :(得分：61)
让我们说你的字母数组如下：“ABCDEFGH”。你有三个索引（i，j，k）表示你将用于当前单词的字母，你可以从：
开始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然后再改变k。
A B C D E F G H ^ ^ ^ i j k A B C D E F G H ^ ^ ^ i j k
一旦你到达G，你也开始改变我。
A B C D E F G H ^ ^ ^ i j k A B C D E F G H ^ ^ ^ i j k ...
用代码编写，看起来像那样
void print_combinations(const char *string)
{
int i, j, k;
int len = strlen(string);
for (i = 0; i < len - 2; i++)
{
for (j = i + 1; j < len - 1; j++)
{
for (k = j + 1; k < len; k++)
printf("%c%c%c\n", string[i], string[j], string[k]);
}
}
}
答案 5 :(得分：52)
以下递归算法从有序集中选择所有k元素组合：
- 选择合并的第一个元素
i - 将
i与大于k-1的元素集合中递归选择的i元素的每个元素组合。
为集合中的每个i迭代上述内容。
必须选择大于i的其余元素，以避免重复。这种方式[3,5]只会被选择一次，因为[3]与[5]结合，而不是两次（条件消除[5] + [3]）。没有这种情况，你会得到变化而不是组合。
答案 6 :(得分：24)
我发现这个线程很有用，并且我认为我会添加一个Javascript解决方案，你可以将其弹出到Firebug中。根据您的JS引擎，如果起始字符串很大，可能需要一些时间。
function string_recurse(active, rest) {
if (rest.length == 0) {
console.log(active);
} else {
string_recurse(active + rest.charAt(0), rest.substring(1, rest.length));
string_recurse(active, rest.substring(1, rest.length));
}
}
string_recurse("", "abc");
输出应如下：
abc
ab
ac
a
bc
b
c
答案 7 :(得分：23)
在C ++中，以下例程将生成[first，last]范围之间的长度距离（first，k）的所有组合：
#include <algorithm>
template <typename Iterator>
bool next_combination(const Iterator first, Iterator k, const Iterator last)
{
/* Credits: Mark Nelson http://marknelson.us */
if ((first == last) || (first == k) || (last == k))
return false;
Iterator i1 = first;
Iterator i2 = last;
++i1;
if (last == i1)
return false;
i1 = last;
--i1;
i1 = k;
--i2;
while (first != i1)
{
if (*--i1 < *i2)
{
Iterator j = k;
while (!(*i1 < *j)) ++j;
std::iter_swap(i1,j);
++i1;
++j;
i2 = k;
std::rotate(i1,j,last);
while (last != j)
{
++j;
++i2;
}
std::rotate(k,i2,last);
return true;
}
}
std::rotate(first,k,last);
return false;
}
可以像这样使用：
#include <string>
#include <iostream>
int main()
{
std::string s = "12345";
std::size_t comb_size = 3;
do
{
std::cout << std::string(s.begin(), s.begin() + comb_size) << std::endl;
} while (next_combination(s.begin(), s.begin() + comb_size, s.end()));
return 0;
}
这将打印以下内容：
123
124
125
134
135
145
234
235
245
345
答案 8 :(得分：20)
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;
}
}
答案 9 :(得分：19)
Python中的简短示例：
def comb(sofar, rest, n):
if n == 0:
print sofar
else:
for i in range(len(rest)):
comb(sofar + rest[i], rest[i+1:], n-1)
>>> comb("", "abcde", 3)
abc
abd
abe
acd
ace
ade
bcd
bce
bde
cde
为了便于说明，使用以下示例描述递归方法：
示例：A B C D E
3的所有组合将是：
- A与其余的所有组合（B C D E）
- B，其余的全部组合为2（C D E）
- C，其余的所有组合（D E）
答案 10 :(得分：17)
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。它产生一个结果，这是一个空列表（即包含空列表的列表）。
对于n＆gt; 0，x遍历列表中的每个元素，xs是x之后的每个元素。
rest使用n - 1的递归调用从xs中选择combinations个元素。函数的最终结果是一个列表，其中每个元素都是x : rest（即x为头部且rest为尾部的列表），用于x的每个不同值和rest。
> combinations 3 "abcde"
["abc","abd","abe","acd","ace","ade","bcd","bce","bde","cde"]
当然，由于Haskell很懒，所以列表会根据需要逐步生成，因此您可以部分评估指数级大组合。
> let c = combinations 8 "abcdefghijklmnopqrstuvwxyz"
> take 10 c
["abcdefgh","abcdefgi","abcdefgj","abcdefgk","abcdefgl","abcdefgm","abcdefgn",
"abcdefgo","abcdefgp","abcdefgq"]
答案 11 :(得分：12)
这就是爷爷COBOL，这是一种备受诽谤的语言。
让我们假设一个包含34个元素的数组，每个元素8个字节（纯粹是任意选择。）这个想法是枚举所有可能的4元素组合并将它们加载到数组中。
我们使用4个指数，每个位置对应4个组中的每个位置
数组的处理方式如下：
idx1 = 1
idx2 = 2
idx3 = 3
idx4 = 4
我们将idx4从4改为结尾。对于每个idx4，我们得到一个独特的组合 四人小组。当idx4到达数组的末尾时，我们将idx3递增1并将idx4设置为idx3 + 1。然后我们再次运行idx4到最后。我们以这种方式进行，分别扩充idx3，idx2和idx1，直到idx1的位置从数组的末尾开始小于4。这完成了算法。
1 --- pos.1
2 --- pos 2
3 --- pos 3
4 --- pos 4
5
6
7
etc.
第一次迭代：
1234
1235
1236
1237
1245
1246
1247
1256
1257
1267
etc.
COBOL示例：
01 DATA_ARAY.
05 FILLER PIC X(8) VALUE "VALUE_01".
05 FILLER PIC X(8) VALUE "VALUE_02".
etc.
01 ARAY_DATA OCCURS 34.
05 ARAY_ITEM PIC X(8).
01 OUTPUT_ARAY OCCURS 50000 PIC X(32).
01 MAX_NUM PIC 99 COMP VALUE 34.
01 INDEXXES COMP.
05 IDX1 PIC 99.
05 IDX2 PIC 99.
05 IDX3 PIC 99.
05 IDX4 PIC 99.
05 OUT_IDX PIC 9(9).
01 WHERE_TO_STOP_SEARCH PIC 99 COMP.
* Stop the search when IDX1 is on the third last array element:
COMPUTE WHERE_TO_STOP_SEARCH = MAX_VALUE - 3
MOVE 1 TO IDX1
PERFORM UNTIL IDX1 > WHERE_TO_STOP_SEARCH
COMPUTE IDX2 = IDX1 + 1
PERFORM UNTIL IDX2 > MAX_NUM
COMPUTE IDX3 = IDX2 + 1
PERFORM UNTIL IDX3 > MAX_NUM
COMPUTE IDX4 = IDX3 + 1
PERFORM UNTIL IDX4 > MAX_NUM
ADD 1 TO OUT_IDX
STRING ARAY_ITEM(IDX1)
ARAY_ITEM(IDX2)
ARAY_ITEM(IDX3)
ARAY_ITEM(IDX4)
INTO OUTPUT_ARAY(OUT_IDX)
ADD 1 TO IDX4
END-PERFORM
ADD 1 TO IDX3
END-PERFORM
ADD 1 TO IDX2
END_PERFORM
ADD 1 TO IDX1
END-PERFORM.
答案 12 :(得分：9)
这是Scala中优雅的通用实现，如99 Scala Problems所述。
object P26 {
def flatMapSublists[A,B](ls: List[A])(f: (List[A]) => List[B]): List[B] =
ls match {
case Nil => Nil
case sublist@(_ :: tail) => f(sublist) ::: flatMapSublists(tail)(f)
}
def combinations[A](n: Int, ls: List[A]): List[List[A]] =
if (n == 0) List(Nil)
else flatMapSublists(ls) { sl =>
combinations(n - 1, sl.tail) map {sl.head :: _}
}
}
答案 13 :(得分：9)
如果您可以使用SQL语法 - 例如，如果您使用LINQ访问结构或数组的字段，或者直接访问具有名为“Alphabet”的表的数据库，只有一个字段“Letter”，则可以适应以下代码：
SELECT A.Letter, B.Letter, C.Letter
FROM Alphabet AS A, Alphabet AS B, Alphabet AS C
WHERE A.Letter<>B.Letter AND A.Letter<>C.Letter AND B.Letter<>C.Letter
AND A.Letter<B.Letter AND B.Letter<C.Letter
这将返回3个字母的所有组合，尽管您在表格“Alphabet”中有多少个字母（可以是3个，8个，10个，27个等）。
如果您想要的是所有排列，而不是组合（即您希望“ACB”和“ABC”计算为不同，而不是只出现一次）只需删除最后一行（AND一个）并完成。
后期编辑：重新阅读问题后，我意识到需要的是通用算法，而不仅仅是选择3个项目的特定算法。亚当休斯的答案是完整的，不幸的是我不能投票（还）。这个答案很简单，但只适用于你想要的3个项目。
答案 14 :(得分：6)
另一个具有懒惰生成组合索引的C＃版本。此版本维护单个索引数组，以定义所有值列表与当前组合值之间的映射，即在整个运行时期间不断使用 O（k）附加空间。该代码在 O（k）时间内生成单独的组合，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
答案 15 :(得分：6)
这里有一个用C＃编码的算法的惰性评估版本：
static bool nextCombination(int[] num, int n, int k)
{
bool finished, changed;
changed = finished = false;
if (k > 0)
{
for (int i = k - 1; !finished && !changed; i--)
{
if (num[i] < (n - 1) - (k - 1) + i)
{
num[i]++;
if (i < k - 1)
{
for (int j = i + 1; j < k; j++)
{
num[j] = num[j - 1] + 1;
}
}
changed = true;
}
finished = (i == 0);
}
}
return changed;
}
static IEnumerable Combinations<T>(IEnumerable<T> elements, int k)
{
T[] elem = elements.ToArray();
int size = elem.Length;
if (k <= size)
{
int[] numbers = new int[k];
for (int i = 0; i < k; i++)
{
numbers[i] = i;
}
do
{
yield return numbers.Select(n => elem[n]);
}
while (nextCombination(numbers, size, k));
}
}
测试部分：
static void Main(string[] args)
{
int k = 3;
var t = new[] { "dog", "cat", "mouse", "zebra"};
foreach (IEnumerable<string> i in Combinations(t, k))
{
Console.WriteLine(string.Join(",", i));
}
}
希望这对你有帮助！
答案 16 :(得分：6)
我有一个用于项目euler的排列算法，在python中：
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
如果
n<len(l)
你应该拥有所需的所有组合而不重复，你需要吗？
它是一个生成器，所以你可以使用它：
for comb in permutation_gen(3,list("ABCDEFGH")):
print comb
答案 17 :(得分：5)
让我们说你的一系列字母看起来像这样：＆＃34; ABCDEFGH＆＃34;。你有三个索引（i，j，k）表示你将用于当前单词的字母，你可以从：
开始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然后再改变k。
A B C D E F G H ^ ^ ^ i j k A B C D E F G H ^ ^ ^ i j k
一旦你到达G，你也开始改变我。
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。
答案 18 :(得分：5)
Clojure版本：
(defn comb [k l]
(if (= 1 k) (map vector l)
(apply concat
(map-indexed
#(map (fn [x] (conj x %2))
(comb (dec k) (drop (inc %1) l)))
l))))
答案 19 :(得分：5)
https://gist.github.com/3118596
JavaScript有一个实现。它具有获取k组合和任何对象数组的所有组合的功能。例子：
k_combinations([1,2,3], 2)
-> [[1,2], [1,3], [2,3]]
combinations([1,2,3])
-> [[1],[2],[3],[1,2],[1,3],[2,3],[1,2,3]]
答案 20 :(得分：5)
Array.prototype.combs = function(num) {
var str = this,
length = str.length,
of = Math.pow(2, length) - 1,
out, combinations = [];
while(of) {
out = [];
for(var i = 0, y; i < length; i++) {
y = (1 << i);
if(y & of && (y !== of))
out.push(str[i]);
}
if (out.length >= num) {
combinations.push(out);
}
of--;
}
return combinations;
}
答案 21 :(得分：3)
这是我最近在Java中编写的代码，它计算并返回“outOf”元素中“num”元素的所有组合。
// author: Sourabh Bhat (heySourabh@gmail.com)
public class Testing
{
public static void main(String[] args)
{
// Test case num = 5, outOf = 8.
int num = 5;
int outOf = 8;
int[][] combinations = getCombinations(num, outOf);
for (int i = 0; i < combinations.length; i++)
{
for (int j = 0; j < combinations[i].length; j++)
{
System.out.print(combinations[i][j] + " ");
}
System.out.println();
}
}
private static int[][] getCombinations(int num, int outOf)
{
int possibilities = get_nCr(outOf, num);
int[][] combinations = new int[possibilities][num];
int arrayPointer = 0;
int[] counter = new int[num];
for (int i = 0; i < num; i++)
{
counter[i] = i;
}
breakLoop: while (true)
{
// Initializing part
for (int i = 1; i < num; i++)
{
if (counter[i] >= outOf - (num - 1 - i))
counter[i] = counter[i - 1] + 1;
}
// Testing part
for (int i = 0; i < num; i++)
{
if (counter[i] < outOf)
{
continue;
} else
{
break breakLoop;
}
}
// Innermost part
combinations[arrayPointer] = counter.clone();
arrayPointer++;
// Incrementing part
counter[num - 1]++;
for (int i = num - 1; i >= 1; i--)
{
if (counter[i] >= outOf - (num - 1 - i))
counter[i - 1]++;
}
}
return combinations;
}
private static int get_nCr(int n, int r)
{
if(r > n)
{
throw new ArithmeticException("r is greater then n");
}
long numerator = 1;
long denominator = 1;
for (int i = n; i >= r + 1; i--)
{
numerator *= i;
}
for (int i = 2; i <= n - r; i++)
{
denominator *= i;
}
return (int) (numerator / denominator);
}
}
答案 22 :(得分：3)
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);
答案 23 :(得分：3)
所有人都说过并完成了O'caml代码。 从代码中可以看出算法很明显。
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 []
;;
答案 24 :(得分：3)
这是我在C ++中的命题
我试图对迭代器类型施加尽可能少的限制，因此这个解决方案假定只是前向迭代器，它可以是一个const_iterator。这适用于任何标准容器。如果参数没有意义，它会抛出std :: invalid_argumnent
#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));
}
}
答案 25 :(得分：3)
我为此在SQL Server 2005中创建了一个解决方案，并将其发布在我的网站上：http://www.jessemclain.com/downloads/code/sql/fn_GetMChooseNCombos.sql.htm
以下是显示用法的示例：
SELECT * FROM dbo.fn_GetMChooseNCombos('ABCD', 2, '')
结果：
Word
----
AB
AC
AD
BC
BD
CD
(6 row(s) affected)
答案 26 :(得分：3)
这是一个方法，它为您提供随机长度字符串中指定大小的所有组合。与quinmars的解决方案类似，但适用于各种输入和k。
代码可以更改为环绕，即输入'abcd'的'dab'w k = 3.
public void run(String data, int howMany){
choose(data, howMany, new StringBuffer(), 0);
}
//n choose k
private void choose(String data, int k, StringBuffer result, int startIndex){
if (result.length()==k){
System.out.println(result.toString());
return;
}
for (int i=startIndex; i<data.length(); i++){
result.append(data.charAt(i));
choose(data,k,result, i+1);
result.setLength(result.length()-1);
}
}
“abcde”的输出：
abc abd abe acd ace ade bcd bce bde cde
答案 27 :(得分：2)
我编写了一个类来处理使用二项式系数的常用函数，这是您的问题所处的问题类型。它执行以下任务：
-
以任意N选择K到文件的格式输出所有K索引。 K索引可以用更具描述性的字符串或字母代替。这种方法使解决这类问题变得非常简单。
-
将K索引转换为已排序二项系数表中条目的正确索引。这种技术比依赖迭代的旧发布技术快得多。它通过使用Pascal三角形中固有的数学属性来实现。我的论文谈到了这一点。我相信我是第一个发现和发布这种技术的人，但我可能错了。
-
将已排序的二项系数表中的索引转换为相应的K索引。
-
使用Mark Dominus方法计算二项式系数，这样就不太可能溢出并使用更大的数字。
-
该类是用.NET C＃编写的，它提供了一种通过使用通用列表来管理与问题相关的对象（如果有）的方法。此类的构造函数采用名为InitTable的bool值，当为true时，将创建一个通用列表来保存要管理的对象。如果此值为false，则不会创建表。不需要创建表来执行上述4种方法。提供访问者方法来访问该表。
-
有一个关联的测试类，它显示了如何使用该类及其方法。它已经过2个案例的广泛测试，并且没有已知的错误。
要阅读此课程并下载代码，请参阅Tablizing The Binomial Coeffieicent。
将此类转换为C ++应该不难。
答案 28 :(得分：2)
算法：
- 从1到2 ^ n。 计数
- 将每个数字转换为二进制表示。
- 翻译每个＆＃39;根据位置指向您的集合中的元素。
在C＃中：
void Main()
{
var set = new [] {"A", "B", "C", "D" }; //, "E", "F", "G", "H", "I", "J" };
var kElement = 2;
for(var i = 1; i < Math.Pow(2, set.Length); i++) {
var result = Convert.ToString(i, 2).PadLeft(set.Length, '0');
var cnt = Regex.Matches(Regex.Escape(result), "1").Count;
if (cnt == kElement) {
for(int j = 0; j < set.Length; j++)
if ( Char.GetNumericValue(result[j]) == 1)
Console.Write(set[j]);
Console.WriteLine();
}
}
}
为什么会这样？
在n元素集和n比特序列的子集之间存在双射。
这意味着我们可以通过计算序列来计算出有多少个子集。
例如，下面的四个元素集可以用{0,1} X {0,1} X {0,1} X {0,1}（或2 ^ 4）个不同的序列来表示。所以 - 所有我们要做的就是从1到2 ^ n计算以找到所有组合。（我们忽略空集。）接下来，将数字转换为它们的二进制表示。然后用你的集合中的元素替换＆＃39; on＆＃39;位。
如果您只想要k个元素结果，则仅在k位为“＃”时打印。
（如果您想要所有子集而不是k长度子集，请删除cnt / kElement部分。）
（有关证明，请参阅麻省理工学院免费课件计算机科学数学，Lehman等，第11.2.2节。https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/readings/）
答案 29 :(得分：1)
void combine(char a[], int N, int M, int m, int start, char result[]) {
if (0 == m) {
for (int i = M - 1; i >= 0; i--)
std::cout << result[i];
std::cout << std::endl;
return;
}
for (int i = start; i < (N - m + 1); i++) {
result[m - 1] = a[i];
combine(a, N, M, m-1, i+1, result);
}
}
void combine(char a[], int N, int M) {
char *result = new char[M];
combine(a, N, M, M, 0, result);
delete[] result;
}
在第一个函数中，m表示你需要选择多少个，而start表示你必须从数组中的哪个位置开始选择。
答案 30 :(得分：1)
JavaScript版本短（ES 5）
spacy_de = spacy.load('de')
spacy_en = spacy.load('en')
def tokenize_de(text):
"""
Tokenizes German text from a string into a list of strings (tokens) and reverses it
"""
return [tok.text for tok in spacy_de.tokenizer(text)][::-1]
def tokenize_en(text):
"""
Tokenizes English text from a string into a list of strings (tokens)
"""
return [tok.text for tok in spacy_en.tokenizer(text)]
SRC = ReversibleField(tokenize = tokenize_de,
init_token = '<sos>',
eos_token = '<eos>',
lower = True,
batch_first= True)
TRG = ReversibleField(tokenize = tokenize_en,
init_token = '<sos>',
eos_token = '<eos>',
lower = True,
batch_first= True)
train_data, valid_data, test_data = Multi30k.splits(exts = ('.de', '.en'),
fields = (SRC, TRG))
SRC.build_vocab(train_data, min_freq = 2)
TRG.build_vocab(train_data, min_freq = 2)
train_iterator, valid_iterator, test_iterator = BucketIterator.splits(
(train_data, valid_data, test_data),
batch_size = BATCH_SIZE,
device = device)
batch = next(iter(train_iterator))
TRG.reverse(batch.trg)
output >>>
['agroupofkidsplayingwithtires.',
'sevenconstructionworkersworkingonabuilding.',
'amanisperformingwithfiresticksbeforeacrowdoutside.']
答案 31 :(得分：1)
C＃的另一个解决方案：
static List<List<T>> GetCombinations<T>(List<T> originalItems, int combinationLength)
{
if (combinationLength < 1)
{
return null;
}
return CreateCombinations<T>(new List<T>(), 0, combinationLength, originalItems);
}
static List<List<T>> CreateCombinations<T>(List<T> initialCombination, int startIndex, int length, List<T> originalItems)
{
List<List<T>> combinations = new List<List<T>>();
for (int i = startIndex; i < originalItems.Count - length + 1; i++)
{
List<T> newCombination = new List<T>(initialCombination);
newCombination.Add(originalItems[i]);
if (length > 1)
{
List<List<T>> newCombinations = CreateCombinations(newCombination, i + 1, length - 1, originalItems);
combinations.AddRange(newCombinations);
}
else
{
combinations.Add(newCombination);
}
}
return combinations;
}
使用示例：
List<char> initialArray = new List<char>() { 'a','b','c','d'};
int combinationLength = 3;
List<List<char>> combinations = GetCombinations(initialArray, combinationLength);
答案 32 :(得分：1)
我正在为PHP寻找类似的解决方案，并且遇到了以下
class Combinations implements Iterator
{
protected $c = null;
protected $s = null;
protected $n = 0;
protected $k = 0;
protected $pos = 0;
function __construct($s, $k) {
if(is_array($s)) {
$this->s = array_values($s);
$this->n = count($this->s);
} else {
$this->s = (string) $s;
$this->n = strlen($this->s);
}
$this->k = $k;
$this->rewind();
}
function key() {
return $this->pos;
}
function current() {
$r = array();
for($i = 0; $i < $this->k; $i++)
$r[] = $this->s[$this->c[$i]];
return is_array($this->s) ? $r : implode('', $r);
}
function next() {
if($this->_next())
$this->pos++;
else
$this->pos = -1;
}
function rewind() {
$this->c = range(0, $this->k);
$this->pos = 0;
}
function valid() {
return $this->pos >= 0;
}
protected function _next() {
$i = $this->k - 1;
while ($i >= 0 && $this->c[$i] == $this->n - $this->k + $i)
$i--;
if($i < 0)
return false;
$this->c[$i]++;
while($i++ < $this->k - 1)
$this->c[$i] = $this->c[$i - 1] + 1;
return true;
}
}
foreach(new Combinations("1234567", 5) as $substring)
echo $substring, ' ';
我不确定课程的效率如何，但我只是将它用于播种机。
答案 33 :(得分：1)
跳上潮流，并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，(List<T> list)是可供选择的列表。返回组合(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;
}
答案 34 :(得分：1)
The Art of Computer Programming, Volume 4A: Combinatorial Algorithms, Part 1 的第7.2.1.3节中的算法L（词典组合）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;
}
答案 35 :(得分：1)
这是一些打印所有C（n，m）组合的简单代码。它的工作原理是初始化并移动一组指向下一个有效组合的数组索引。初始化指数以指向最低的m个指数（按字典顺序表示最小的组合）。然后，从第m个索引开始，我们尝试向前移动索引。如果索引已达到其限制，我们将尝试上一个索引（一直到索引1）。如果我们可以向前移动指数，那么我们将重置所有更大的指数。
m=(rand()%n)+1; // m will vary from 1 to n
for (i=0;i<n;i++) a[i]=i+1;
// we want to print all possible C(n,m) combinations of selecting m objects out of n
printf("Printing C(%d,%d) possible combinations ...\n", n,m);
// This is an adhoc algo that keeps m pointers to the next valid combination
for (i=0;i<m;i++) p[i]=i; // the p[.] contain indices to the a vector whose elements constitute next combination
done=false;
while (!done)
{
// print combination
for (i=0;i<m;i++) printf("%2d ", a[p[i]]);
printf("\n");
// update combination
// method: start with p[m-1]. try to increment it. if it is already at the end, then try moving p[m-2] ahead.
// if this is possible, then reset p[m-1] to 1 more than (the new) p[m-2].
// if p[m-2] can not also be moved, then try p[m-3]. move that ahead. then reset p[m-2] and p[m-1].
// repeat all the way down to p[0]. if p[0] can not also be moved, then we have generated all combinations.
j=m-1;
i=1;
move_found=false;
while ((j>=0) && !move_found)
{
if (p[j]<(n-i))
{
move_found=true;
p[j]++; // point p[j] to next index
for (k=j+1;k<m;k++)
{
p[k]=p[j]+(k-j);
}
}
else
{
j--;
i++;
}
}
if (!move_found) done=true;
}
答案 36 :(得分：1)
短python代码，产生索引位置
def yield_combos(n,k):
# n is set size, k is combo size
i = 0
a = [0 for i in range(k)]
while i > -1:
for j in range(i+1, k):
a[j] = a[j-1]+1
i=j
yield a
while a[i] == i + n - k:
i -= 1
a[i] += 1
答案 37 :(得分：1)
简短的PHP算法，基于java解决方案从n（二项式系数）返回k个元素的所有组合：
$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);
答案 38 :(得分：1)
Lisp宏生成所有值r（一次采用）
的代码(defmacro txaat (some-list taken-at-a-time)
(let* ((vars (reverse (truncate-list '(a b c d e f g h i j) taken-at-a-time))))
`(
,@(loop for i below taken-at-a-time
for j in vars
with nested = nil
finally (return nested)
do
(setf
nested
`(loop for ,j from
,(if (< i (1- (length vars)))
`(1+ ,(nth (1+ i) vars))
0)
below (- (length ,some-list) ,i)
,@(if (equal i 0)
`(collect
(list
,@(loop for k from (1- taken-at-a-time) downto 0
append `((nth ,(nth k vars) ,some-list)))))
`(append ,nested))))))))
所以，
CL-USER> (macroexpand-1 '(txaat '(a b c d) 1))
(LOOP FOR A FROM 0 TO (- (LENGTH '(A B C D)) 1)
COLLECT (LIST (NTH A '(A B C D))))
T
CL-USER> (macroexpand-1 '(txaat '(a b c d) 2))
(LOOP FOR A FROM 0 TO (- (LENGTH '(A B C D)) 2)
APPEND (LOOP FOR B FROM (1+ A) TO (- (LENGTH '(A B C D)) 1)
COLLECT (LIST (NTH A '(A B C D)) (NTH B '(A B C D)))))
T
CL-USER> (macroexpand-1 '(txaat '(a b c d) 3))
(LOOP FOR A FROM 0 TO (- (LENGTH '(A B C D)) 3)
APPEND (LOOP FOR B FROM (1+ A) TO (- (LENGTH '(A B C D)) 2)
APPEND (LOOP FOR C FROM (1+ B) TO (- (LENGTH '(A B C D)) 1)
COLLECT (LIST (NTH A '(A B C D))
(NTH B '(A B C D))
(NTH C '(A B C D))))))
T
CL-USER>
和
CL-USER> (txaat '(a b c d) 1)
((A) (B) (C) (D))
CL-USER> (txaat '(a b c d) 2)
((A B) (A C) (A D) (B C) (B D) (C D))
CL-USER> (txaat '(a b c d) 3)
((A B C) (A B D) (A C D) (B C D))
CL-USER> (txaat '(a b c d) 4)
((A B C D))
CL-USER> (txaat '(a b c d) 5)
NIL
CL-USER> (txaat '(a b c d) 0)
NIL
CL-USER>
答案 39 :(得分：1)
#include <stdio.h>
unsigned int next_combination(unsigned int *ar, size_t n, unsigned int k)
{
unsigned int finished = 0;
unsigned int changed = 0;
unsigned int i;
if (k > 0) {
for (i = k - 1; !finished && !changed; i--) {
if (ar[i] < (n - 1) - (k - 1) + i) {
/* Increment this element */
ar[i]++;
if (i < k - 1) {
/* Turn the elements after it into a linear sequence */
unsigned int j;
for (j = i + 1; j < k; j++) {
ar[j] = ar[j - 1] + 1;
}
}
changed = 1;
}
finished = i == 0;
}
if (!changed) {
/* Reset to first combination */
for (i = 0; i < k; i++) {
ar[i] = i;
}
}
}
return changed;
}
typedef void(*printfn)(const void *, FILE *);
void print_set(const unsigned int *ar, size_t len, const void **elements,
const char *brackets, printfn print, FILE *fptr)
{
unsigned int i;
fputc(brackets[0], fptr);
for (i = 0; i < len; i++) {
print(elements[ar[i]], fptr);
if (i < len - 1) {
fputs(", ", fptr);
}
}
fputc(brackets[1], fptr);
}
int main(void)
{
unsigned int numbers[] = { 0, 1, 2 };
char *elements[] = { "a", "b", "c", "d", "e" };
const unsigned int k = sizeof(numbers) / sizeof(unsigned int);
const unsigned int n = sizeof(elements) / sizeof(const char*);
do {
print_set(numbers, k, (void*)elements, "[]", (printfn)fputs, stdout);
putchar('\n');
} while (next_combination(numbers, n, k));
getchar();
return 0;
}
答案 40 :(得分：1)
由于未提及编程语言，我假设列表也可以。所以这是一个适合短列表的OCaml版本（非尾递归）。如果列出 l 任何类型的元素和一个整数 n ，它将会如果我们假设元素的顺序，则返回包含 n l 元素的所有可能列表的列表结果列表被忽略，即列表['a';'b']与['b';'a']相同，并将报告一次。因此结果列表的大小将是（（List.length l）选择n）。
递归的直觉如下：你取出列表的头部然后进行两次递归调用：
- 递归调用1（RC1）：到列表的尾部，但是选择n-1个元素
- 递归调用2（RC2）：到列表的尾部，但是选择n个元素
将递归结果，list-multiply（请带有奇数名称）列表的头部与RC1的结果相结合，然后追加（@）RC2的结果。 List-multiply是以下操作lmul：
a lmul [ l1 ; l2 ; l3] = [a::l1 ; a::l2 ; a::l3]
lmul在下面的代码中实现为
List.map (fun x -> h::x)
当列表的大小等于您要选择的元素数时，终止递归，在这种情况下，您只需返回列表本身。
所以这是OCaml中的四线程，它实现了上述算法：
let rec choose l n = match l, (List.length l) with
| _, lsize when n==lsize -> [l]
| h::t, _ -> (List.map (fun x-> h::x) (choose t (n-1))) @ (choose t n)
| [], _ -> []
答案 41 :(得分：1)
这是一个 Clojure 版本，它使用我在 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)))))))))
它提供了三种调用它的方法：
（a）对于 n 所选项目，正如问题所要求的那样：
user=> (count (select 3 "abcdefgh"))
56
（b）：
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))
user=> (select '(1 2 3))
(() (3) (2) (1) (2 3) (1 3) (1 2) (1 2 3))
答案 42 :(得分：1)
这是一个递归程序，它假定集合中nCk.Elements的组合从1到n
#include<stdio.h>
#include<stdlib.h>
int nCk(int n,int loopno,int ini,int *a,int k)
{
static int count=0;
int i;
loopno--;
if(loopno<0)
{
a[k-1]=ini;
for(i=0;i<k;i++)
{
printf("%d,",a[i]);
}
printf("\n");
count++;
return 0;
}
for(i=ini;i<=n-loopno-1;i++)
{
a[k-1-loopno]=i+1;
nCk(n,loopno,i+1,a,k);
}
if(ini==0)
return count;
else
return 0;
}
void main()
{
int n,k,*a,count;
printf("Enter the value of n and k\n");
scanf("%d %d",&n,&k);
a=(int*)malloc(k*sizeof(int));
count=nCk(n,k,0,a,k);
printf("No of combinations=%d\n",count);
}
答案 43 :(得分：1)
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)
答案 44 :(得分：0)
另一个python递归解决方案。
def combination_indicies(n, k, j = 0, stack = []):
if len(stack) == k:
yield list(stack)
return
for i in range(j, n):
stack.append(i)
for x in combination_indicies(n, k, i + 1, stack):
yield x
stack.pop()
list(combination_indicies(5, 3))
输出：
[[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]]
答案 45 :(得分：0)
这是一个简单且难以理解的递归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
}
答案 46 :(得分：0)
最近在 IronScripter 网站上有一个 PowerShell 挑战需要一个 n-choose-k 解决方案。我在那里发布了一个解决方案，但这里有一个更通用的版本。
- AllK 开关用于控制输出是只是长度 ChooseK 的组合，还是长度 1 到 ChooseK 的组合。
- Prefix 参数实际上是输出字符串的累加器，但其作用是为初始调用传入的值实际上将作为每行输出的前缀。
function Get-NChooseK
{
[CmdletBinding()]
Param
(
[String[]]
$ArrayN
, [Int]
$ChooseK
, [Switch]
$AllK
, [String]
$Prefix = ''
)
PROCESS
{
# Validate the inputs
$ArrayN = $ArrayN | Sort-Object -Unique
If ($ChooseK -gt $ArrayN.Length)
{
Write-Error "Can't choose $ChooseK items when only $($ArrayN.Length) are available." -ErrorAction Stop
}
# Control the output
$firstK = If ($AllK) { 1 } Else { $ChooseK }
# Get combinations
$firstK..$ChooseK | ForEach-Object {
$thisK = $_
$ArrayN[0..($ArrayN.Length-($thisK--))] | ForEach-Object {
If ($thisK -eq 0)
{
Write-Output ($Prefix+$_)
}
Else
{
Get-NChooseK -Array ($ArrayN[($ArrayN.IndexOf($_)+1)..($ArrayN.Length-1)]) -Choose $thisK -AllK:$false -Prefix ($Prefix+$_)
}
}
}
}
}
例如：
PS C:\>$ArrayN = 'E','B','C','A','D'
PS C:\>$ChooseK = 3
PS C:\>Get-NChooseK -ArrayN $ArrayN -ChooseK $ChooseK
ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE
答案 47 :(得分：0)
以下是C ++中的一种算法，不使用，既不使用STL，也不使用递归或条件嵌套循环。这样更快，它不执行任何元素交换，并且不给递归带来负担，并且可以通过替换mallloc()，free()和{{1}轻松地移植到ANSIC。 }分别用于printf()，new和delete。
如果要显示字母不同或更长的元素，请更改std::cout参数以指向与*alphabet不同的字符串。
"abcdefg"重要提示：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 - 1) { //Exit condition (all of the values in the array are flush to the left)
delete[] ka;
return;
}
ki++;
}
}
}
int main(int argc, char *argv[])
{
GenCombinations(7, 4, "abcdefg");
return 0;
}
参数必须指向至少包含*alphabet个字符的字符串。您还可以传递在其他地方定义的字符串的地址。
组合：超出“ 7选择4”。
答案 48 :(得分：0)
这是我在javascript中的贡献（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>")
}
答案 49 :(得分：0)
我知道已经有很多答案，但是我想我会在JavaScript中添加自己的个人贡献，其中包括两个函数-一个用于生成一个变量的所有可能的不同k子集。原始n元素集，并使用第一个函数生成原始n元素集的幂集。
这是两个函数的代码：
var animation = new MatrixAnimationUsingPath
{
PathGeometry = animationPath,
Duration = TimeSpan.FromSeconds(5)
};
var transform = new MatrixTransform();
child.RenderTransform = transform;
child.Loaded += (s, e) =>
transform.BeginAnimation(MatrixTransform.MatrixProperty, animation);
我使用集合[“ a”，“ b”，“ c”，“ d”，“ e”，“ f”]作为基础集对此进行了测试，并运行了代码以产生以下功率集：
//Generate combination subsets from a base set of elements (passed as an array). This function should generate an
//array containing nCr elements, where nCr = n!/[r! (n-r)!].
//Arguments:
//[1] baseSet : The base set to create the subsets from (e.g., ["a", "b", "c", "d", "e", "f"])
//[2] cnt : The number of elements each subset is to contain (e.g., 3)
function MakeCombinationSubsets(baseSet, cnt)
{
var bLen = baseSet.length;
var indices = [];
var subSet = [];
var done = false;
var result = []; //Contains all the combination subsets generated
var done = false;
var i = 0;
var idx = 0;
var tmpIdx = 0;
var incr = 0;
var test = 0;
var newIndex = 0;
var inBounds = false;
var tmpIndices = [];
var checkBounds = false;
//First, generate an array whose elements are indices into the base set ...
for (i=0; i<cnt; i++)
indices.push(i);
//Now create a clone of this array, to be used in the loop itself ...
tmpIndices = [];
tmpIndices = tmpIndices.concat(indices);
//Now initialise the loop ...
idx = cnt - 1; //point to the last element of the indices array
incr = 0;
done = false;
while (!done)
{
//Create the current subset ...
subSet = []; //Make sure we begin with a completely empty subset before continuing ...
for (i=0; i<cnt; i++)
subSet.push(baseSet[tmpIndices[i]]); //Create the current subset, using items selected from the
//base set, using the indices array (which will change as we
//continue scanning) ...
//Add the subset thus created to the result set ...
result.push(subSet);
//Now update the indices used to select the elements of the subset. At the start, idx will point to the
//rightmost index in the indices array, but the moment that index moves out of bounds with respect to the
//base set, attention will be shifted to the next left index.
test = tmpIndices[idx] + 1;
if (test >= bLen)
{
//Here, we're about to move out of bounds with respect to the base set. We therefore need to scan back,
//and update indices to the left of the current one. Find the leftmost index in the indices array that
//isn't going to move out of bounds with respect to the base set ...
tmpIdx = idx - 1;
incr = 1;
inBounds = false; //Assume at start that the index we're checking in the loop below is out of bounds
checkBounds = true;
while (checkBounds)
{
if (tmpIdx < 0)
{
checkBounds = false; //Exit immediately at this point
}
else
{
newIndex = tmpIndices[tmpIdx] + 1;
test = newIndex + incr;
if (test >= bLen)
{
//Here, incrementing the current selected index will take that index out of bounds, so
//we move on to the next index to the left ...
tmpIdx--;
incr++;
}
else
{
//Here, the index will remain in bounds if we increment it, so we
//exit the loop and signal that we're in bounds ...
inBounds = true;
checkBounds = false;
//End if/else
}
//End if
}
//End while
}
//At this point, if we'er still in bounds, then we continue generating subsets, but if not, we abort immediately.
if (!inBounds)
done = true;
else
{
//Here, we're still in bounds. We need to update the indices accordingly. NOTE: at this point, although a
//left positioned index in the indices array may still be in bounds, incrementing it to generate indices to
//the right may take those indices out of bounds. We therefore need to check this as we perform the index
//updating of the indices array.
tmpIndices[tmpIdx] = newIndex;
inBounds = true;
checking = true;
i = tmpIdx + 1;
while (checking)
{
test = tmpIndices[i - 1] + 1; //Find out if incrementing the left adjacent index takes it out of bounds
if (test >= bLen)
{
inBounds = false; //If we move out of bounds, exit NOW ...
checking = false;
}
else
{
tmpIndices[i] = test; //Otherwise, update the indices array ...
i++; //Now move on to the next index to the right in the indices array ...
checking = (i < cnt); //And continue until we've exhausted all the indices array elements ...
//End if/else
}
//End while
}
//At this point, if the above updating of the indices array has moved any of its elements out of bounds,
//we abort subset construction from this point ...
if (!inBounds)
done = true;
//End if/else
}
}
else
{
//Here, the rightmost index under consideration isn't moving out of bounds with respect to the base set when
//we increment it, so we simply increment and continue the loop ...
tmpIndices[idx] = test;
//End if
}
//End while
}
return(result);
//End function
}
function MakePowerSet(baseSet)
{
var bLen = baseSet.length;
var result = [];
var i = 0;
var partialSet = [];
result.push([]); //add the empty set to the power set
for (i=1; i<bLen; i++)
{
partialSet = MakeCombinationSubsets(baseSet, i);
result = result.concat(partialSet);
//End i loop
}
//Now, finally, add the base set itself to the power set to make it complete ...
partialSet = [];
partialSet.push(baseSet);
result = result.concat(partialSet);
return(result);
//End function
}
只需按原样复制和粘贴这两个函数，您将具备提取n个元素集的不同k子集，并所需的基本能力所需的基础知识。如果需要，可以设置该n个元素。
我并不是说这很优雅，只是它经过大量测试（并且在调试阶段变成了蓝色：））。
答案 50 :(得分：0)
JavaScript，基于生成器的递归方法：
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()">
>= k:<input id="kinp" oninput="test()"> >= 1
<div id="log"></div>
通过这种方式（减小i和unshift()），它会以降序生成组合和组合中的元素，使眼睛有些悦目。
测试在1秒钟后停止，因此输入怪异数字相对安全。
答案 51 :(得分：0)
PowerShell 解决方案：
function Get-NChooseK
{
<#
.SYNOPSIS
Returns all the possible combinations by choosing K items at a time from N possible items.
.DESCRIPTION
Returns all the possible combinations by choosing K items at a time from N possible items.
The combinations returned do not consider the order of items as important i.e. 123 is considered to be the same combination as 231, etc.
.PARAMETER ArrayN
The array of items to choose from.
.PARAMETER ChooseK
The number of items to choose.
.PARAMETER AllK
Includes combinations for all lesser values of K above zero i.e. 1 to K.
.PARAMETER Prefix
String that will prefix each line of the output.
.EXAMPLE
PS C:\> Get-NChooseK -ArrayN '1','2','3' -ChooseK 3
123
.EXAMPLE
PS C:\> Get-NChooseK -ArrayN '1','2','3' -ChooseK 3 -AllK
1
2
3
12
13
23
123
.EXAMPLE
PS C:\> Get-NChooseK -ArrayN '1','2','3' -ChooseK 2 -Prefix 'Combo: '
Combo: 12
Combo: 13
Combo: 23
.NOTES
Author : nmbell
#>
# Use cmdlet binding
[CmdletBinding()]
# Declare parameters
Param
(
[String[]]
$ArrayN
, [Int]
$ChooseK
, [Switch]
$AllK
, [String]
$Prefix = ''
)
BEGIN
{
}
PROCESS
{
# Validate the inputs
$ArrayN = $ArrayN | Sort-Object -Unique
If ($ChooseK -gt $ArrayN.Length)
{
Write-Error "Can't choose $ChooseK items when only $($ArrayN.Length) are available." -ErrorAction Stop
}
# Control the output
$firstK = If ($AllK) { 1 } Else { $ChooseK }
# Get combinations
$firstK..$ChooseK | ForEach-Object {
$thisK = $_
$ArrayN[0..($ArrayN.Length-($thisK--))] | ForEach-Object {
If ($thisK -eq 0)
{
Write-Output ($Prefix+$_)
}
Else
{
Get-NChooseK -Array ($ArrayN[($ArrayN.IndexOf($_)+1)..($ArrayN.Length-1)]) -Choose $thisK -AllK:$false -Prefix ($Prefix+$_)
}
}
}
}
END
{
}
}
例如：
PS C:\>Get-NChooseK -ArrayN 'A','B','C','D','E' -ChooseK 3
ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE
最近在 IronScripter 网站上发布了一个与此问题类似的挑战，您可以在其中找到指向我的和其他一些解决方案的链接。
答案 52 :(得分：0)
下面的Haskell代码可同时计算组合数和组合，由于Haskell的惰性，您可以在不计算另一部分的情况下得到其中的一部分。
import Data.Semigroup
import Data.Monoid
data Comb = MkComb {count :: Int, combinations :: [[Int]]} deriving (Show, Eq, Ord)
instance Semigroup Comb where
(MkComb c1 cs1) <> (MkComb c2 cs2) = MkComb (c1 + c2) (cs1 ++ cs2)
instance Monoid Comb where
mempty = MkComb 0 []
addElem :: Comb -> Int -> Comb
addElem (MkComb c cs) x = MkComb c (map (x :) cs)
comb :: Int -> Int -> Comb
comb n k | n < 0 || k < 0 = error "error in `comb n k`, n and k should be natural number"
comb n k | k == 0 || k == n = MkComb 1 [(take k [k-1,k-2..0])]
comb n k | n < k = mempty
comb n k = comb (n-1) k <> (comb (n-1) (k-1) `addElem` (n-1))
它的工作原理是：
*Main> comb 0 1
MkComb {count = 0, combinations = []}
*Main> comb 0 0
MkComb {count = 1, combinations = [[]]}
*Main> comb 1 1
MkComb {count = 1, combinations = [[0]]}
*Main> comb 4 2
MkComb {count = 6, combinations = [[1,0],[2,0],[2,1],[3,0],[3,1],[3,2]]}
*Main> count (comb 10 5)
252
答案 53 :(得分：0)
这里是使用宏的Lisp方法。这可在Common Lisp中使用，并应在其他Lisp方言中使用。
下面的代码创建“ n”个嵌套循环，并为列表body中“ n”个元素的每种组合执行任意代码块（存储在lst变量中）。变量var指向包含用于循环的变量的列表。
(defmacro do-combinations ((var lst num) &body body)
(loop with syms = (loop repeat num collect (gensym))
for i on syms
for k = `(loop for ,(car i) on (cdr ,(cadr i))
do (let ((,var (list ,@(reverse syms)))) (progn ,@body)))
then `(loop for ,(car i) on ,(if (cadr i) `(cdr ,(cadr i)) lst) do ,k)
finally (return k)))
让我们看看...
(macroexpand-1 '(do-combinations (p '(1 2 3 4 5 6 7) 4) (pprint (mapcar #'car p))))
(LOOP FOR #:G3217 ON '(1 2 3 4 5 6 7) DO
(LOOP FOR #:G3216 ON (CDR #:G3217) DO
(LOOP FOR #:G3215 ON (CDR #:G3216) DO
(LOOP FOR #:G3214 ON (CDR #:G3215) DO
(LET ((P (LIST #:G3217 #:G3216 #:G3215 #:G3214)))
(PROGN (PPRINT (MAPCAR #'CAR P))))))))
(do-combinations (p '(1 2 3 4 5 6 7) 4) (pprint (mapcar #'car p)))
(1 2 3 4)
(1 2 3 5)
(1 2 3 6)
...
由于默认情况下不存储组合，因此将存储量降至最低。选择body代码而不是存储所有结果的可能性也提供了更大的灵活性。
答案 54 :(得分：0)
在Python中，利用递归和一切都通过引用完成的事实。对于非常大的集合，这将占用大量内存，但具有初始集合可以是复杂对象的优点。它只能找到独特的组合。
import copy
def find_combinations( length, set, combinations = None, candidate = None ):
# recursive function to calculate all unique combinations of unique values
# from [set], given combinations of [length]. The result is populated
# into the 'combinations' list.
#
if combinations == None:
combinations = []
if candidate == None:
candidate = []
for item in set:
if item in candidate:
# this item already appears in the current combination somewhere.
# skip it
continue
attempt = copy.deepcopy(candidate)
attempt.append(item)
# sorting the subset is what gives us completely unique combinations,
# so that [1, 2, 3] and [1, 3, 2] will be treated as equals
attempt.sort()
if len(attempt) < length:
# the current attempt at finding a new combination is still too
# short, so add another item to the end of the set
# yay recursion!
find_combinations( length, set, combinations, attempt )
else:
# the current combination attempt is the right length. If it
# already appears in the list of found combinations then we'll
# skip it.
if attempt in combinations:
continue
else:
# otherwise, we append it to the list of found combinations
# and move on.
combinations.append(attempt)
continue
return len(combinations)
你这样使用它。传递'结果'是可选的，所以你可以用它来获得可能组合的数量......虽然这样效率很低（最好通过计算完成）。
size = 3
set = [1, 2, 3, 4, 5]
result = []
num = find_combinations( size, set, result )
print "size %d results in %d sets" % (size, num)
print "result: %s" % (result,)
您应该从该测试数据中获得以下输出：
size 3 results in 10 sets
result: [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5]]
如果您的设置如下所示，它也会起作用：
set = [
[ 'vanilla', 'cupcake' ],
[ 'chocolate', 'pudding' ],
[ 'vanilla', 'pudding' ],
[ 'chocolate', 'cookie' ],
[ 'mint', 'cookie' ]
]
答案 55 :(得分：0)
这是一个简单的JS解决方案：
function getAllCombinations(n, k, f1) {
indexes = Array(k);
for (let i =0; i< k; i++) {
indexes[i] = i;
}
var total = 1;
f1(indexes);
while (indexes[0] !== n-k) {
total++;
getNext(n, indexes);
f1(indexes);
}
return {total};
}
function getNext(n, vec) {
const k = vec.length;
vec[k-1]++;
for (var i=0; i<k; i++) {
var currentIndex = k-i-1;
if (vec[currentIndex] === n - i) {
var nextIndex = k-i-2;
vec[nextIndex]++;
vec[currentIndex] = vec[nextIndex] + 1;
}
}
for (var i=1; i<k; i++) {
if (vec[i] === n - (k-i - 1)) {
vec[i] = vec[i-1] + 1;
}
}
return vec;
}
let start = new Date();
let result = getAllCombinations(10, 3, indexes => console.log(indexes));
let runTime = new Date() - start;
console.log({
result, runTime
});
答案 56 :(得分：0)
作为迭代器对象实现的MetaTrader MQL4的非常快速的组合。
代码很容易理解。
我对很多算法进行了基准测试，这个算法非常快 - 比大多数next_combination（）函数快3倍。
＆＃13;
＆＃13;
＆＃13;
＆＃13;
class CombinationsIterator
{
private:
int input_array[]; // 1 2 3 4 5
int index_array[]; // i j k
int m_elements; // N
int m_indices; // K
public:
CombinationsIterator(int &src_data[], int k)
{
m_indices = k;
m_elements = ArraySize(src_data);
ArrayCopy(input_array, src_data);
ArrayResize(index_array, m_indices);
// create initial combination (0..k-1)
for (int i = 0; i < m_indices; i++)
{
index_array[i] = i;
}
}
// https://stackoverflow.com/questions/5076695
// bool next_combination(int &item[], int k, int N)
bool advance()
{
int N = m_elements;
for (int i = m_indices - 1; i >= 0; --i)
{
if (index_array[i] < --N)
{
++index_array[i];
for (int j = i + 1; j < m_indices; ++j)
{
index_array[j] = index_array[j - 1] + 1;
}
return true;
}
}
return false;
}
void getItems(int &items[])
{
// fill items[] from input array
for (int i = 0; i < m_indices; i++)
{
items[i] = input_array[index_array[i]];
}
}
};
用于测试上述迭代器类的驱动程序：
＆＃13;
＆＃13;
＆＃13;
＆＃13;
//+------------------------------------------------------------------+
//| |
//+------------------------------------------------------------------+
// driver program to test above class
#define N 5
#define K 3
void OnStart()
{
int myset[N] = {1, 2, 3, 4, 5};
int items[K];
CombinationsIterator comboIt(myset, K);
do
{
comboIt.getItems(items);
printf("%s", ArrayToString(items));
} while (comboIt.advance());
}
＆＃13;
＆＃13;
＆＃13;
＆＃13;
Output:
1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5
答案 57 :(得分：0)
我为C ++中的组合创建了一个通用类。 就像这样使用。
char ar[] = "0ABCDEFGH";
nCr ncr(8, 3);
while(ncr.next()) {
for(int i=0; i<ncr.size(); i++) cout << ar[ncr[i]];
cout << ' ';
}
我的图书馆ncr [i]从1而不是从0返回。 这就是为什么数组中有0的原因。 如果你想考虑订单，只需将nCr类变为nPr。 用法完全相同。
结果
ABC ABD ABE ABF ABG ABH ACD 高手 ACF ACG ACH ADE ADF ADG ADH AEF AEG AEH AFG AFH AGH BCD BCE BCF BCG BCH BDE BDF BDG BDH BEF 求 BEH BFG BFH BGH CDE CDF CDG CDH CEF CEG CEH CFG CFH CGH DEF DEG DEH DFG DFH DGH EFG EFH EGH FGH
这是头文件。
#pragma once
#include <exception>
class NRexception : public std::exception
{
public:
virtual const char* what() const throw() {
return "Combination : N, R should be positive integer!!";
}
};
class Combination
{
public:
Combination(int n, int r);
virtual ~Combination() { delete [] ar;}
int& operator[](unsigned i) {return ar[i];}
bool next();
int size() {return r;}
static int factorial(int n);
protected:
int* ar;
int n, r;
};
class nCr : public Combination
{
public:
nCr(int n, int r);
bool next();
int count() const;
};
class nTr : public Combination
{
public:
nTr(int n, int r);
bool next();
int count() const;
};
class nHr : public nTr
{
public:
nHr(int n, int r) : nTr(n,r) {}
bool next();
int count() const;
};
class nPr : public Combination
{
public:
nPr(int n, int r);
virtual ~nPr() {delete [] on;}
bool next();
void rewind();
int count() const;
private:
bool* on;
void inc_ar(int i);
};
并实施。
#include "combi.h"
#include <set>
#include<cmath>
Combination::Combination(int n, int r)
{
//if(n < 1 || r < 1) throw NRexception();
ar = new int[r];
this->n = n;
this->r = r;
}
int Combination::factorial(int n)
{
return n == 1 ? n : n * factorial(n-1);
}
int nPr::count() const
{
return factorial(n)/factorial(n-r);
}
int nCr::count() const
{
return factorial(n)/factorial(n-r)/factorial(r);
}
int nTr::count() const
{
return pow(n, r);
}
int nHr::count() const
{
return factorial(n+r-1)/factorial(n-1)/factorial(r);
}
nCr::nCr(int n, int r) : Combination(n, r)
{
if(r == 0) return;
for(int i=0; i<r-1; i++) ar[i] = i + 1;
ar[r-1] = r-1;
}
nTr::nTr(int n, int r) : Combination(n, r)
{
for(int i=0; i<r-1; i++) ar[i] = 1;
ar[r-1] = 0;
}
bool nCr::next()
{
if(r == 0) return false;
ar[r-1]++;
int i = r-1;
while(ar[i] == n-r+2+i) {
if(--i == -1) return false;
ar[i]++;
}
while(i < r-1) ar[i+1] = ar[i++] + 1;
return true;
}
bool nTr::next()
{
ar[r-1]++;
int i = r-1;
while(ar[i] == n+1) {
ar[i] = 1;
if(--i == -1) return false;
ar[i]++;
}
return true;
}
bool nHr::next()
{
ar[r-1]++;
int i = r-1;
while(ar[i] == n+1) {
if(--i == -1) return false;
ar[i]++;
}
while(i < r-1) ar[i+1] = ar[i++];
return true;
}
nPr::nPr(int n, int r) : Combination(n, r)
{
on = new bool[n+2];
for(int i=0; i<n+2; i++) on[i] = false;
for(int i=0; i<r; i++) {
ar[i] = i + 1;
on[i] = true;
}
ar[r-1] = 0;
}
void nPr::rewind()
{
for(int i=0; i<r; i++) {
ar[i] = i + 1;
on[i] = true;
}
ar[r-1] = 0;
}
bool nPr::next()
{
inc_ar(r-1);
int i = r-1;
while(ar[i] == n+1) {
if(--i == -1) return false;
inc_ar(i);
}
while(i < r-1) {
ar[++i] = 0;
inc_ar(i);
}
return true;
}
void nPr::inc_ar(int i)
{
on[ar[i]] = false;
while(on[++ar[i]]);
if(ar[i] != n+1) on[ar[i]] = true;
}
答案 58 :(得分：0)
我们可以使用位的概念来做到这一点。我们有一个＆＃34; abc，＆＃34;我们希望所有元素的组合长度为2（即＆＃34; ab＆＃34;，＆＃34; ac＆＃34;，＆＃34; bc＆＃34;。）
我们可以找到数字范围从1到2 ^ n（不包括）的设置位。这里1到7，无论我们在哪里设置bits = 2，我们都可以从字符串中打印相应的值。
例如：
- 1 - 001
- 2 - 010
- 3 - 011 - ＆gt;
print ab (str[0] , str[1]) - 4 - 100
- 5 - 101 - ＆gt;
print ac (str[0] , str[2]) - 6 - 110 - ＆gt;
print ab (str[1] , str[2]) - 7 - 111。
代码示例：
public class StringCombinationK {
static void combk(String s , int k){
int n = s.length();
int num = 1<<n;
int j=0;
int count=0;
for(int i=0;i<num;i++){
if (countSet(i)==k){
setBits(i,j,s);
count++;
System.out.println();
}
}
System.out.println(count);
}
static void setBits(int i,int j,String s){ // print the corresponding string value,j represent the index of set bit
if(i==0){
return;
}
if(i%2==1){
System.out.print(s.charAt(j));
}
setBits(i/2,j+1,s);
}
static int countSet(int i){ //count number of set bits
if( i==0){
return 0;
}
return (i%2==0? 0:1) + countSet(i/2);
}
public static void main(String[] arhs){
String s = "abcdefgh";
int k=3;
combk(s,k);
}
}
答案 59 :(得分：0)
您可以使用 Asif 算法生成所有可能的组合。这可能是最简单、最有效的方法。您可以查看中等文章here。
让我们看看在 JavaScript 中的实现。
function Combinations( arr, r ) {
// To avoid object referencing, cloning the array.
arr = arr && arr.slice() || [];
var len = arr.length;
if( !len || r > len || !r )
return [ [] ];
else if( r === len )
return [ arr ];
if( r === len ) return arr.reduce( ( x, v ) => {
x.push( [ v ] );
return x;
}, [] );
var head = arr.shift();
return Combinations( arr, r - 1 ).map( x => {
x.unshift( head );
return x;
} ).concat( Combinations( arr, r ) );
}
// Now do your stuff.
console.log( Combinations( [ 'a', 'b', 'c', 'd', 'e' ], 3 ) );
答案 60 :(得分：0)
这是我的JavaScript解决方案，通过使用reduce / map更加实用，几乎消除了所有变量
＆＃13;
＆＃13;
＆＃13;
＆＃13;
function combinations(arr, size) {
var len = arr.length;
if (size > len) return [];
if (!size) return [[]];
if (size == len) return [arr];
return arr.reduce(function (acc, val, i) {
var res = combinations(arr.slice(i + 1), size - 1)
.map(function (comb) { return [val].concat(comb); });
return acc.concat(res);
}, []);
}
var combs = combinations([1,2,3,4,5,6,7,8],3);
combs.map(function (comb) {
document.body.innerHTML += comb.toString() + '<br />';
});
document.body.innerHTML += '<br /> Total combinations = ' + combs.length;
答案 61 :(得分：0)
也许我错过了这一点（你需要算法，而不是现成的解决方案），但似乎scala开箱即用（现在）：
def combis(str:String, k:Int):Array[String] = {
str.combinations(k).toArray
}
使用这样的方法：
println(combis("abcd",2).toList)
将产生：
List(ab, ac, ad, bc, bd, cd)
答案 62 :(得分：0)
快速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;
}
答案 63 :(得分：0)
像Andrea Ambu这样的Python，但没有硬编码选择三个。
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
答案 64 :(得分：0)
这是一个C ++解决方案，我想出了使用递归和位移。它也可以在C中工作。
void r_nCr(unsigned int startNum, unsigned int bitVal, unsigned int testNum) // Should be called with arguments (2^r)-1, 2^(r-1), 2^(n-1)
{
unsigned int n = (startNum - bitVal) << 1;
n += bitVal ? 1 : 0;
for (unsigned int i = log2(testNum) + 1; i > 0; i--) // Prints combination as a series of 1s and 0s
cout << (n >> (i - 1) & 1);
cout << endl;
if (!(n & testNum) && n != startNum)
r_nCr(n, bitVal, testNum);
if (bitVal && bitVal < testNum)
r_nCr(startNum, bitVal >> 1, testNum);
}
您可以找到有关其工作原理的解释here。
答案 65 :(得分：0)
C＃简单算法。 （我发布它是因为我试图使用你们上传的那个，但由于某种原因我无法编译它 - 扩展一个类？所以我写了我自己的一个以防万一有人否则我遇到了同样的问题）。 顺便说一下，我对c＃的关注度不比基本编程多，但是这个工作正常。
public static List<List<int>> GetSubsetsOfSizeK(List<int> lInputSet, int k)
{
List<List<int>> lSubsets = new List<List<int>>();
GetSubsetsOfSizeK_rec(lInputSet, k, 0, new List<int>(), lSubsets);
return lSubsets;
}
public static void GetSubsetsOfSizeK_rec(List<int> lInputSet, int k, int i, List<int> lCurrSet, List<List<int>> lSubsets)
{
if (lCurrSet.Count == k)
{
lSubsets.Add(lCurrSet);
return;
}
if (i >= lInputSet.Count)
return;
List<int> lWith = new List<int>(lCurrSet);
List<int> lWithout = new List<int>(lCurrSet);
lWith.Add(lInputSet[i++]);
GetSubsetsOfSizeK_rec(lInputSet, k, i, lWith, lSubsets);
GetSubsetsOfSizeK_rec(lInputSet, k, i, lWithout, lSubsets);
}
用法：GetSubsetsOfSizeK(set of type List<int>, integer k)
您可以修改它以迭代您正在使用的任何内容。
祝你好运！答案 66 :(得分：0)
这是我提出的解决这个问题的算法。它用c ++编写，但几乎可以适用于任何支持按位操作的语言。
void r_nCr(const unsigned int &startNum, const unsigned int &bitVal, const unsigned int &testNum) // Should be called with arguments (2^r)-1, 2^(r-1), 2^(n-1)
{
unsigned int n = (startNum - bitVal) << 1;
n += bitVal ? 1 : 0;
for (unsigned int i = log2(testNum) + 1; i > 0; i--) // Prints combination as a series of 1s and 0s
cout << (n >> (i - 1) & 1);
cout << endl;
if (!(n & testNum) && n != startNum)
r_nCr(n, bitVal, testNum);
if (bitVal && bitVal < testNum)
r_nCr(startNum, bitVal >> 1, testNum);
}
您可以看到有关其工作原理的解释here。
答案 67 :(得分：0)
这是一个coffeescript实现
combinations: (list, n) ->
permuations = Math.pow(2, list.length) - 1
out = []
combinations = []
while permuations
out = []
for i in [0..list.length]
y = ( 1 << i )
if( y & permuations and (y isnt permuations))
out.push(list[i])
if out.length <= n and out.length > 0
combinations.push(out)
permuations--
return combinations
答案 68 :(得分：0)
另一个递归解决方案（你应该能够使用字母而不是数字来移植它）使用堆栈，比大多数情况下要短一些：
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选择3或我想要从0到4
开始的所有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]
答案 69 :(得分：0)
短快速C实现
#include <stdio.h>
void main(int argc, char *argv[]) {
const int n = 6; /* The size of the set; for {1, 2, 3, 4} it's 4 */
const int p = 4; /* The size of the subsets; for {1, 2}, {1, 3}, ... it's 2 */
int comb[40] = {0}; /* comb[i] is the index of the i-th element in the combination */
int i = 0;
for (int j = 0; j <= n; j++) comb[j] = 0;
while (i >= 0) {
if (comb[i] < n + i - p + 1) {
comb[i]++;
if (i == p - 1) { for (int j = 0; j < p; j++) printf("%d ", comb[j]); printf("\n"); }
else { comb[++i] = comb[i - 1]; }
} else i--; }
}
要查看速度有多快，请使用此代码并对其进行测试
#include <time.h>
#include <stdio.h>
void main(int argc, char *argv[]) {
const int n = 32; /* The size of the set; for {1, 2, 3, 4} it's 4 */
const int p = 16; /* The size of the subsets; for {1, 2}, {1, 3}, ... it's 2 */
int comb[40] = {0}; /* comb[i] is the index of the i-th element in the combination */
int c = 0; int i = 0;
for (int j = 0; j <= n; j++) comb[j] = 0;
while (i >= 0) {
if (comb[i] < n + i - p + 1) {
comb[i]++;
/* if (i == p - 1) { for (int j = 0; j < p; j++) printf("%d ", comb[j]); printf("\n"); } */
if (i == p - 1) c++;
else { comb[++i] = comb[i - 1]; }
} else i--; }
printf("%d!%d == %d combination(s) in %15.3f second(s)\n ", p, n, c, clock()/1000.0);
}
使用cmd.exe（windows）进行测试：
Microsoft Windows XP [Version 5.1.2600]
(C) Copyright 1985-2001 Microsoft Corp.
c:\Program Files\lcc\projects>combination
16!32 == 601080390 combination(s) in 5.781 second(s)
c:\Program Files\lcc\projects>
度过愉快的一天。
答案 70 :(得分：0)
递归地，一个非常简单的答案，combo，在Free Pascal中。
procedure combinata (n, k :integer; producer :oneintproc);
procedure combo (ndx, nbr, len, lnd :integer);
begin
for nbr := nbr to len do begin
productarray[ndx] := nbr;
if len < lnd then
combo(ndx+1,nbr+1,len+1,lnd)
else
producer(k);
end;
end;
begin
combo (0, 0, n-k, n-1);
end;
＆＃34;生产者＆＃34;处理为每种组合制作的产品阵列。
答案 71 :(得分：0)
不需要收集操作。问题几乎与循环K嵌套循环相同，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。
答案 72 :(得分：0)
我想提出我的解决方案。 next中没有递归调用，也没有嵌套循环。
代码的核心是next()方法。
public class Combinations {
final int pos[];
final List<Object> set;
public Combinations(List<?> l, int k) {
pos = new int[k];
set=new ArrayList<Object>(l);
reset();
}
public void reset() {
for (int i=0; i < pos.length; ++i) pos[i]=i;
}
public boolean next() {
int i = pos.length-1;
for (int maxpos = set.size()-1; pos[i] >= maxpos; --maxpos) {
if (i==0) return false;
--i;
}
++pos[i];
while (++i < pos.length)
pos[i]=pos[i-1]+1;
return true;
}
public void getSelection(List<?> l) {
@SuppressWarnings("unchecked")
List<Object> ll = (List<Object>)l;
if (ll.size()!=pos.length) {
ll.clear();
for (int i=0; i < pos.length; ++i)
ll.add(set.get(pos[i]));
}
else {
for (int i=0; i < pos.length; ++i)
ll.set(i, set.get(pos[i]));
}
}
}
用法示例：
static void main(String[] args) {
List<Character> l = new ArrayList<Character>();
for (int i=0; i < 32; ++i) l.add((char)('a'+i));
Combinations comb = new Combinations(l,5);
int n=0;
do {
++n;
comb.getSelection(l);
//Log.debug("%d: %s", n, l.toString());
} while (comb.next());
Log.debug("num = %d", n);
}
答案 73 :(得分：0)
简单但缓慢的C ++回溯算法。
#include <iostream>
void backtrack(int* numbers, int n, int k, int i, int s)
{
if (i == k)
{
for (int j = 0; j < k; ++j)
{
std::cout << numbers[j];
}
std::cout << std::endl;
return;
}
if (s > n)
{
return;
}
numbers[i] = s;
backtrack(numbers, n, k, i + 1, s + 1);
backtrack(numbers, n, k, i, s + 1);
}
int main(int argc, char* argv[])
{
int n = 5;
int k = 3;
int* numbers = new int[k];
backtrack(numbers, n, k, 0, 1);
delete[] numbers;
return 0;
}
答案 74 :(得分：0)
这个答案怎么样...这打印长度为3的所有组合...它可以推广任何长度... 工作代码......
#include<iostream>
#include<string>
using namespace std;
void combination(string a,string dest){
int l = dest.length();
if(a.empty() && l == 3 ){
cout<<dest<<endl;}
else{
if(!a.empty() && dest.length() < 3 ){
combination(a.substr(1,a.length()),dest+a[0]);}
if(!a.empty() && dest.length() <= 3 ){
combination(a.substr(1,a.length()),dest);}
}
}
int main(){
string demo("abcd");
combination(demo,"");
return 0;
}
答案 75 :(得分：-1)
#include <unistd.h>
#include <stdio.h>
#include <iconv.h>
#include <string.h>
#include <errno.h>
#include <stdlib.h>
int main(int argc, char **argv)
{
int opt = -1, min_len = 0, max_len = 0;
char ofile[256], fchar[2], tchar[2];
ofile[0] = 0;
fchar[0] = 0;
tchar[0] = 0;
while((opt = getopt(argc, argv, "o:f:t:l:L:")) != -1)
{
switch(opt)
{
case 'o':
strncpy(ofile, optarg, 255);
break;
case 'f':
strncpy(fchar, optarg, 1);
break;
case 't':
strncpy(tchar, optarg, 1);
break;
case 'l':
min_len = atoi(optarg);
break;
case 'L':
max_len = atoi(optarg);
break;
default:
printf("usage: %s -oftlL\n\t-o output file\n\t-f from char\n\t-t to char\n\t-l min seq len\n\t-L max seq len", argv[0]);
}
}
if(max_len < 1)
{
printf("error, length must be more than 0\n");
return 1;
}
if(min_len > max_len)
{
printf("error, max length must be greater or equal min_length\n");
return 1;
}
if((int)fchar[0] > (int)tchar[0])
{
printf("error, invalid range specified\n");
return 1;
}
FILE *out = fopen(ofile, "w");
if(!out)
{
printf("failed to open input file with error: %s\n", strerror(errno));
return 1;
}
int cur_len = min_len;
while(cur_len <= max_len)
{
char buf[cur_len];
for(int i = 0; i < cur_len; i++)
buf[i] = fchar[0];
fwrite(buf, cur_len, 1, out);
fwrite("\n", 1, 1, out);
while(buf[0] != (tchar[0]+1))
{
while(buf[cur_len-1] < tchar[0])
{
(int)buf[cur_len-1]++;
fwrite(buf, cur_len, 1, out);
fwrite("\n", 1, 1, out);
}
if(cur_len < 2)
break;
if(buf[0] == tchar[0])
{
bool stop = true;
for(int i = 1; i < cur_len; i++)
{
if(buf[i] != tchar[0])
{
stop = false;
break;
}
}
if(stop)
break;
}
int u = cur_len-2;
for(; u>=0 && buf[u] >= tchar[0]; u--)
;
(int)buf[u]++;
for(int i = u+1; i < cur_len; i++)
buf[i] = fchar[0];
fwrite(buf, cur_len, 1, out);
fwrite("\n", 1, 1, out);
}
cur_len++;
}
fclose(out);
return 0;
}
这里是我在c ++中的实现，它将所有组合写入指定的文件，但行为可以改变，我生成各种字典，它接受最小和最大长度和字符范围，目前只支持ansi，它足够我的需要
答案 76 :(得分：-2)
如果您是C ++用户，您可以利用next_permutation函数：
string name = "abcdefg";
int desiredCharsNumber = 3;
string chosenChars = string(name.size()-desiredCharsNumber,'0') + string(desiredCharsNumber, '1');
do {
for (int i = 0; i < name.size(); ++i)
if (chosenChars[i] == '1')
cout << name[i];
cout << endl;
} while (next_permutation(chosenChars.begin(), chosenChars.end()));
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