我想编写一个函数，它将一组字母作为参数，并选择其中的一些字母。
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。那么你应该得到：
8! / ((8 - 3)! * 3!) = 56
数组（或单词）作为回报，每个包含 3 个字母。
IQue
问问题
522802 次
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445
计算机编程艺术第 4 卷：分册 3有很多这些内容可能比我描述的更适合您的特定情况。
格雷码
您将遇到的一个问题当然是内存，而且很快，您的集合中的 20 个元素就会出现问题 - 20 C 3 = 1140。如果您想遍历集合，最好使用修改后的灰色代码算法，因此您不会将所有这些都保存在内存中。这些从前一个组合生成下一个组合并避免重复。其中有许多用于不同的用途。我们想最大化连续组合之间的差异吗？最小化？等等。
一些描述格雷码的原始论文：
以下是一些涵盖该主题的其他论文：
- Eades, Hickey, Read Adjacent Interchange Combination Generation Algorithm 的有效实现（PDF，Pascal 代码）
- 组合生成器
- 组合格雷码调查(PostScript)
- 格雷码算法
Chase's Twiddle (算法)
Phillip J Chase，“算法 382：N 个对象中 M 个的组合”（1970 年）
C中的算法...
按字典顺序排列的组合索引（Buckles 算法 515）
您还可以通过索引（按字典顺序）引用组合。意识到索引应该是基于索引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，我们有一个集合 {1,2,3,4,5,6}... 我们想要三个元素。假设 {1,2,3} 我们可以说元素之间的差异是一个且有序且最小。{1,2,4} 有一个变化，按字典顺序排列为 2。因此，最后一个位置的“变化”数说明了字典顺序的一个变化。第二位，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正比），所以它解释了更多变化。
我描述的方法是一种解构，看起来，从集合到索引，我们需要做相反的事情——这要复杂得多。这就是Buckles解决问题的方法。我写了一些C 来计算它们，稍作改动——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。笔记：
- 由于组合是无序的，{1,3,2} = {1,2,3}——我们按字典顺序排列它们。
- 此方法有一个隐含的 0 来启动第一个差异的集合。
按字典顺序排列的组合索引 (McCaffrey)
还有另一种方式：它的概念更容易掌握和编程，但它没有 Buckles 的优化。幸运的是，它也不会产生重复的组合：
最大化的集合
，其中
。
举个例子：27 = C(6,4) + C(5,3) + C(2,2) + C(1,1)。所以，第 27 个字典组合的四件事是：{1,2,5,6}，这些是您想要查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
出于教学目的，提供了以下两种算法。他们实现了迭代器和（更通用的）文件夹整体组合。它们尽可能快，复杂度为 O( n C k )。内存消耗受k.
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
更通用的版本将调用用户提供的函数以及状态变量，从初始状态开始。由于我们需要在不同状态之间传递状态，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
于 2008-09-24T15:24:50.300 回答
206
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
于 2009-12-14T03:24:55.617 回答
92
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]
于 2013-04-27T19:36:38.657 回答
82
我可以介绍我对这个问题的递归 Python 解决方案吗？
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))
示例用法：
>>> len(list(choose_iter("abcdefgh",3)))
56
我喜欢它的简单性。
于 2010-05-14T21:15:34.873 回答
70
假设您的字母数组如下所示：“ABCDEFGH”。您有三个索引 (i, j, k) 指示您要为当前单词使用哪些字母，您可以从以下内容开始：
ABCDEFGH ^ ^ ^ 伊克
首先你改变k，所以下一步看起来像这样：
ABCDEFGH ^ ^ ^ 伊克
如果你到达终点，你继续改变 j ，然后再改变 k 。
ABCDEFGH ^ ^ ^ 伊克 ABCDEFGH ^ ^ ^ 伊克
一旦你 j 达到 G，你也开始改变 i。
ABCDEFGH ^ ^ ^ 伊克 ABCDEFGH ^ ^ ^ 伊克 ...
写在代码中，看起来像这样
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]);
}
}
}
于 2008-09-24T15:31:36.757 回答
57
以下递归算法从有序集中挑选所有 k 元素组合：
i选择组合的第一个元素- 与从大于 的元素集合中递归选择的元素组合
i中的每一个组合。k-1i
i对集合中的每个重复上述内容。
必须选择大于 的其余元素i，以避免重复。这样 [3,5] 将只被选择一次，因为 [3] 与 [5] 结合，而不是两次（条件消除了 [5] + [3]）。如果没有这个条件，你会得到变化而不是组合。
于 2008-09-24T17:40:35.713 回答
32
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
为了解释，递归方法用下面的例子来描述：
示例：ABCDE
3 的所有组合将是：
- A 与其余 2 的所有组合 (BCDE)
- B 与其余 2 的所有组合 (CDE)
- C 与其余 2 的所有组合 (DE)
于 2013-08-01T14:27:35.970 回答
26
在 C++ 中，以下例程将生成范围 [first,last) 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;
}
它可以这样使用：
#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
于 2009-10-24T11:58:21.873 回答
26
我发现这个线程很有用，并认为我会添加一个可以弹出到 Firebug 的 Javascript 解决方案。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");
输出应如下所示：
abc
ab
ac
a
bc
b
c
于 2011-11-17T17:42:48.387 回答
21
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;
}
}
于 2008-09-24T15:37:26.807 回答
20
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，即一个空列表（即包含一个空列表的列表）。
对于 n > 0，x遍历列表的每个元素并且xs是 之后的每个元素x。
rest使用对 的递归调用来挑选n - 1元素。该函数的最终结果是一个列表，其中每个元素都是（即具有作为头部和作为尾部的列表）对于每个不同的值和。xscombinationsx : restxrestxrest
> combinations 3 "abcde"
["abc","abd","abe","acd","ace","ade","bcd","bce","bde","cde"]
当然，由于 Haskell 是惰性的，列表会根据需要逐渐生成，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"]
于 2011-12-24T17:49:38.670 回答
14
COBOL 出现了，这是一种备受诟病的语言。
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一个数组中。
我们使用 4 个索引，每个索引对应 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 变化到末尾。对于每个 idx4，我们得到一个独特的四组组合。当 idx4 到达数组末尾时，我们将 idx3 加 1 并将 idx4 设置为 idx3+1。然后我们再次运行 idx4 到最后。我们以这种方式继续，分别增加 idx3、idx2 和 idx1，直到 idx1 的位置距离数组末尾小于 4。这样就完成了算法。
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.
于 2012-11-21T09:47:28.207 回答
10
另一个具有延迟生成组合索引的 C# 版本。这个版本维护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
于 2015-05-13T21:20:38.380 回答
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”的表且只有一个字符字段“Letter”的数据库，则可以调整以下内容代码：
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，无论您在表“Alphabet”中有多少个字母（可以是 3、8、10、27 等）。
如果您想要的是所有排列，而不是组合（即您希望“ACB”和“ABC”算作不同，而不是只出现一次），只需删除最后一行（AND 行）就可以了。
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。亚当休斯的回答是完整的，不幸的是我不能投票（还）。这个答案很简单，但仅适用于您想要 3 个项目的情况。
于 2008-09-24T16:59:58.683 回答
9
这是 Scala 中一个优雅的通用实现，如99 Scala Problems中所述。
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 :: _}
}
}
于 2010-04-08T19:30:36.460 回答
8
我有一个用于项目 euler 的置换算法，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)
你应该有你需要的所有组合而不重复，你需要吗？
它是一个生成器，所以你可以像这样使用它：
for comb in permutation_gen(3,list("ABCDEFGH")):
print comb
于 2008-09-25T07:13:00.783 回答
7
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]]
于 2012-07-15T21:05:57.150 回答
6
在这里，您有一个用 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));
}
}
和测试部分：
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！
于 2010-12-26T18:43:05.860 回答
6
假设您的字母数组如下所示：“ABCDEFGH”。您有三个索引 (i, j, k) 指示您要为当前单词使用哪些字母，您可以从以下内容开始：
ABCDEFGH ^ ^ ^ 伊克
首先你改变k，所以下一步看起来像这样：
ABCDEFGH ^ ^ ^ 伊克
如果你到达终点，你继续改变 j ，然后再改变 k 。
ABCDEFGH ^ ^ ^ 伊克 ABCDEFGH ^ ^ ^ 伊克
一旦你 j 达到 G，你也开始改变 i。
ABCDEFGH ^ ^ ^ 伊克 ABCDEFGH ^ ^ ^ 伊克 ...
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。
于 2017-05-18T00:04:21.193 回答
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;
}
于 2012-05-21T18:53:51.963 回答
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))))
于 2014-01-04T03:56:13.673 回答
4
这是一种从随机长度字符串中为您提供指定大小的所有组合的方法。类似于 quinmars 的解决方案，但适用于不同的输入和 k。
可以将代码更改为环绕，即输入'abcd' wk=3 中的'dab'。
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
于 2011-12-13T20:20:33.953 回答
4
这里所说的和所做的都是 O'caml 代码。从代码中可以看出算法..
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 []
;;
于 2014-03-03T13:52:52.970 回答
4
算法：
- 从 1 数到 2^n。
- 将每个数字转换为其二进制表示。
- 根据位置将每个“开启”位转换为集合的元素。
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();
}
}
}
为什么它有效？
在 n 元素集合的子集和 n 位序列之间存在双射。
这意味着我们可以通过计算序列来计算出有多少子集。
例如，下面的四个元素集合可以用{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 -我们所要做的就是从 1 数到 2^n 来找到所有的组合。（我们忽略空集。）接下来，将数字转换为它们的二进制表示。然后将您的集合中的元素替换为“on”位。
如果您只需要 k 个元素结果，则仅在 k 位为“on”时打印。
（如果您想要所有子集而不是 k 个长度的子集，请删除 cnt/kElement 部分。）
（有关证明，请参阅 MIT 免费课件计算机科学数学，Lehman 等人，第 11.2.2 节。https: //ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics- for-computer-science-fall-2010/readings/ )
于 2017-02-12T17:38:54.173 回答
3
这是我在 C++ 中的提议
我试图对迭代器类型施加尽可能少的限制，所以这个解决方案假设只是前向迭代器，它可以是一个 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));
}
}
于 2008-09-25T07:00:54.120 回答
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
这是一个显示用法的示例：
SELECT * FROM dbo.fn_GetMChooseNCombos('ABCD', 2, '')
结果：
Word
----
AB
AC
AD
BC
BD
CD
(6 row(s) affected)
于 2009-06-30T14:48:02.657 回答
3
这是我最近用 Java 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);
}
}
于 2010-03-13T13:10:35.313 回答
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);
于 2014-09-26T12:52:15.157 回答
3
简短的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]*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
于 2015-04-18T10:21:56.683 回答
3
简短的 javascript 版本 (ES 5)
let combine = (list, n) =>
n == 0 ?
[[]] :
list.flatMap((e, i) =>
combine(
list.slice(i + 1),
n - 1
).map(c => [e].concat(c))
);
let res = combine([1,2,3,4], 3);
res.forEach(e => console.log(e.join()));于 2020-09-06T06:47:11.117 回答
2
我编写了一个类来处理处理二项式系数的常用函数，这是您的问题所属的问题类型。它执行以下任务：
将任何 N 选择 K 的所有 K 索引以良好的格式输出到文件。K-indexes 可以替换为更具描述性的字符串或字母。这种方法使得解决这类问题变得非常简单。
将 K 索引转换为已排序二项式系数表中条目的正确索引。这种技术比依赖迭代的旧已发布技术快得多。它通过使用帕斯卡三角形固有的数学属性来做到这一点。我的论文谈到了这一点。我相信我是第一个发现并发表这种技术的人，但我可能是错的。
将已排序二项式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，它不太可能溢出并且适用于更大的数字。
该类是用 .NET C# 编写的，并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一个名为 InitTable 的 bool 值，当它为 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。无需创建表即可执行上述 4 种方法。提供了访问器方法来访问表。
有一个关联的测试类，它显示了如何使用该类及其方法。它已经用 2 个案例进行了广泛的测试，并且没有已知的错误。
要了解此类并下载代码，请参阅制表二项式系数。
将此类转换为 C++ 应该不难。
于 2012-09-25T07:49:04.030 回答
2
基于java解决方案的短php算法从n（二项式系数）返回k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 中：
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);
于 2015-02-03T09:20:41.570 回答
2
这是我的 JavaScript 解决方案，通过使用 reduce/map 功能更强大，它消除了几乎所有变量
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;于 2015-05-16T17:44:19.743 回答
2
The Art of Computer Programming, Volume 4A: Combinatorial Algorithms, Part 1 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;
}
于 2015-12-26T15:14:58.273 回答
2
赶上潮流，并发布另一个解决方案。这是一个通用的 Java 实现。输入：(int k)是要选择的元素数量，是可供选择(List<T> list)的列表。返回组合列表(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;
}
于 2016-01-04T09:48:47.987 回答
2
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>这种方式（递减i和unshift()）以递减的顺序产生组合和组合内的元素，有点令人赏心悦目。
测试在 1 秒后停止，因此输入奇怪的数字相对安全。
于 2019-05-30T16:06:56.027 回答
1
这是一些打印所有 C(n,m) 组合的简单代码。它通过初始化和移动一组指向下一个有效组合的数组索引来工作。索引被初始化为指向最低的 m 个索引（按字典顺序排列的最小组合）。然后，从第 m 个索引开始，我们尝试将索引向前移动。如果索引已达到其限制，我们会尝试前一个索引（一直到索引 1）。如果我们可以向前移动一个索引，那么我们会重置所有更大的索引。
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;
}
于 2010-04-06T09:18:17.670 回答
1
Lisp 宏为所有值 r 生成代码（一次获取）
(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>
于 2011-07-17T15:41:04.860 回答
1
这是一个递归程序，它生成集合的组合，nCk.Elements假设集合是从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);
}
于 2012-01-20T19:26:06.900 回答
1
在 VB.Net 中，该算法从一组数字（PoolArray）中收集 n 个数字的所有组合。例如，来自“8,10,20,33,41,44,47”的 5 个选项的所有组合。
Sub CreateAllCombinationsOfPicksFromPool(ByVal PicksArray() As UInteger, ByVal PicksIndex As UInteger, ByVal PoolArray() As UInteger, ByVal PoolIndex As UInteger)
If PicksIndex < PicksArray.Length Then
For i As Integer = PoolIndex To PoolArray.Length - PicksArray.Length + PicksIndex
PicksArray(PicksIndex) = PoolArray(i)
CreateAllCombinationsOfPicksFromPool(PicksArray, PicksIndex + 1, PoolArray, i + 1)
Next
Else
' completed combination. build your collections using PicksArray.
End If
End Sub
Dim PoolArray() As UInteger = Array.ConvertAll("8,10,20,33,41,44,47".Split(","), Function(u) UInteger.Parse(u))
Dim nPicks as UInteger = 5
Dim Picks(nPicks - 1) As UInteger
CreateAllCombinationsOfPicksFromPool(Picks, 0, PoolArray, 0)
于 2012-01-24T11:41:29.623 回答
1
由于没有提到编程语言，我假设列表也可以。所以这里有一个适合短列表（非尾递归）的 OCaml 版本。给定一个包含任何类型元素的列表l和一个整数n ，如果我们假设结果列表中元素的顺序被忽略，它将返回包含l的n 个元素的所有可能列表的列表，即 list ['a'; 'b'] 与 ['b';'a'] 相同，将报告一次。所以结果列表的大小将是（（List.length l）选择n）。
递归的直觉如下：取列表的头部，然后进行两次递归调用：
- 递归调用 1 (RC1)：到列表的尾部，但选择 n-1 个元素
- 递归调用 2 (RC2)：到列表的尾部，但选择 n 个元素
结合递归结果，列表乘法（请使用奇数名称）列表的头部与 RC1 的结果，然后附加 (@) RC2 的结果。列表乘法是以下操作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，递归终止，在这种情况下，您只需返回列表本身。
所以这里有一个实现上述算法的 OCaml 中的四行代码：
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)
| [], _ -> []
于 2012-03-15T19:05:33.060 回答
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 表示您还需要选择多少，start 表示您必须从数组中的哪个位置开始选择。
于 2012-07-22T19:32:43.847 回答
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)))))))))
它提供了三种调用方式：
(a)对于问题要求的恰好n 个选定项目：
user=> (count (select 3 "abcdefgh"))
56
(b)对于n1和n2之间的选定项目：
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)对于0到集合大小之间的选定项：
user=> (select '(1 2 3))
(() (3) (2) (1) (2 3) (1 3) (1 2) (1 2 3))
于 2013-02-11T21:17:11.633 回答
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)
于 2013-07-04T15:59:57.370 回答
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;
}
于 2013-07-14T12:19:34.560 回答
1
我正在为 PHP 寻找类似的解决方案并遇到以下问题
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, ' ';
我不确定这门课的效率如何，但我只是将它用于播种机。
于 2016-06-08T13:42:05.230 回答
1
使用 C# 的另一种解决方案：
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);
于 2017-02-02T10:41:22.420 回答
1
我们可以使用位的概念来做到这一点。让我们有一个字符串“abc”，我们想要所有长度为 2 的元素组合（即“ab”、“ac”、“bc”。）
我们可以找到从 1 到 2^n（不包括）的数字中的设置位。这里是 1 到 7，无论我们设置 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 ->
print ab (str[0] , str[1]) - 4 - 100
- 5 - 101 ->
print ac (str[0] , str[2]) - 6 - 110 ->
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);
}
}
于 2017-06-27T17:09:38.250 回答
1
这是使用宏的 Lisp 方法。这适用于 Common Lisp 并且应该适用于其他 Lisp 方言。
body下面的代码创建“n”个嵌套循环，并为 list 中“n”个元素的每个组合执行任意代码块（存储在变量中） lst。该变量var指向一个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)))
让我们来看看...
(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，因此将存储量保持在最低限度。选择代码而不是存储所有结果的可能性body也提供了更大的灵活性。
于 2019-03-21T03:45:18.937 回答
1
跟着 Haskell 代码同时计算组合数和组合，由于 Haskell 的懒惰，你可以得到其中一部分而不计算另一部分。
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))
它的工作原理如下：
*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
于 2019-05-15T03:19:42.940 回答
1
我知道已经有很多答案了，但我想我会在 JavaScript 中添加我自己的个人贡献，它由两个函数组成 - 一个用于生成原始 n- 的所有可能不同的 k 子集元素集，以及使用第一个函数生成原始 n 元素集的幂集。
下面是这两个函数的代码：
//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
}
我以集合 ["a", "b", "c", "d", "e", "f"] 作为基集对此进行了测试，并运行代码以生成以下幂集：
[]
["a"]
["b"]
["c"]
["d"]
["e"]
["f"]
["a","b"]
["a","c"]
["a","d"]
["a","e"]
["a","f"]
["b","c"]
["b","d"]
["b","e"]
["b","f"]
["c","d"]
["c","e"]
["c","f"]
["d","e"]
["d","f"]
["e","f"]
["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"]
["a","b","c","d"]
["a","b","c","e"]
["a","b","c","f"]
["a","b","d","e"]
["a","b","d","f"]
["a","b","e","f"]
["a","c","d","e"]
["a","c","d","f"]
["a","c","e","f"]
["a","d","e","f"]
["b","c","d","e"]
["b","c","d","f"]
["b","c","e","f"]
["b","d","e","f"]
["c","d","e","f"]
["a","b","c","d","e"]
["a","b","c","d","f"]
["a","b","c","e","f"]
["a","b","d","e","f"]
["a","c","d","e","f"]
["b","c","d","e","f"]
["a","b","c","d","e","f"]
只需“按原样”复制并粘贴这两个函数，您将拥有提取 n 元素集的不同 k 子集所需的基础知识，并根据需要生成该 n 元素集的幂集。
我并不认为这很优雅，只是在经过大量测试后它可以工作（并在调试阶段变成蓝色:)）。
于 2019-11-30T00:35:22.740 回答
1
另一个python recusive 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]]
于 2021-03-22T21:52:35.267 回答
0
在像 Andrea Ambu 这样的 Python 中，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
于 2008-12-04T00:02:09.517 回答
0
在 Python 中，利用递归和一切都是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。它只会找到唯一的组合。
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)
你用这种方式。传递“结果”是可选的，因此您可以使用它来获取可能组合的数量……尽管这确实效率低下（最好通过计算来完成）。
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]]
如果你的集合看起来像这样，它也能正常工作：
set = [
[ 'vanilla', 'cupcake' ],
[ 'chocolate', 'pudding' ],
[ 'vanilla', 'pudding' ],
[ 'chocolate', 'cookie' ],
[ 'mint', 'cookie' ]
]
于 2011-12-24T16:34:58.143 回答
0
这是我在 javascript 中的贡献（无递归）
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>")
}
于 2012-05-03T20:42:36.120 回答
0
这个答案怎么样......这会打印长度为 3 的所有组合......它可以推广到任何长度......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;
}
于 2013-01-12T09:52:11.507 回答
0
简短的快速 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--; }
}
要查看它有多快，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>
祝你今天过得愉快。
于 2013-04-27T15:33:21.040 回答
0
使用堆栈的另一个递归解决方案（您应该能够将其移植为使用字母而不是数字），虽然比大多数短一点：
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 或者我想要从 0 到 4 开始的所有 3 个数字组合
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]
于 2013-05-12T06:46:26.670 回答
0
这是一个咖啡脚本实现
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
于 2013-06-26T19:07:23.980 回答
0
也许我错过了重点（您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)
于 2014-05-28T11:21:17.283 回答
0
简短的快速 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;
}
于 2014-06-28T13:24:33.843 回答
0
这是我使用递归和位移提出的 C++ 解决方案。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);
}
您可以在此处找到有关其工作原理的说明。
于 2014-12-05T08:23:08.507 回答
0
C# 简单算法。（我发布它是因为我尝试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，以防其他人面临同样的问题我做到了）。顺便说一句，我对 c# 的兴趣并不多于基本编程，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！
于 2015-01-19T19:38:26.520 回答
0
这是我想出的解决这个问题的算法。它是用 c++ 编写的，但几乎可以适应任何支持位运算的语言。
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);
}
您可以在此处查看其工作原理的说明。
于 2015-02-03T19:48:26.587 回答
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;
“生产者”处理为每个组合制作的 productarray。
于 2016-04-25T13:43:12.250 回答
0
不需要集合操作。这个问题几乎与循环 K 嵌套循环相同，但您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。
于 2016-12-14T22:35:37.177 回答
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;
}
于 2017-01-03T13:26:52.877 回答
0
我为 C++ 中的组合制作了一个通用类。它是这样使用的。
char ar[] = "0ABCDEFGH";
nCr ncr(8, 3);
while(ncr.next()) {
for(int i=0; i<ncr.size(); i++) cout << ar[ncr[i]];
cout << ' ';
}
我的库 ncr[i] 从 1 返回，而不是从 0 返回。这就是数组中有 0 的原因。如果要考虑顺序，只需将 nCr 类更改为 nPr。用法相同。
结果
ABC ABD ABE ABF ABG ABH ACD ACE ACF ACG ACH ADE ADF ADG ADH AEF AEG AEH AFG AFH AGH BCD BCE BCF BCG BCH BDE BDF BDG BDH BEF BEG BEH BFG BFH BGH CDE CDF CDG CDH CEF CEG CEH CFG CFH CGH DEF DEG DEH DFG DFH DGH EFG EFH EGH FGH
这里是头文件。
#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);
};
和实施。
#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;
}
于 2018-02-22T00:27:26.943 回答
0
作为迭代器对象实现的 MetaTrader MQL4 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() 函数快大约 3 倍。
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]];
}
}
};测试上述迭代器类的驱动程序：
//+------------------------------------------------------------------+
//| |
//+------------------------------------------------------------------+
// 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());
}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于 2018-02-27T22:43:50.580 回答
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
});于 2018-07-14T15:57:39.323 回答
0
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。这种方式速度更快，它不执行任何元素交换，也不会给堆栈带来递归负担，并且还可以通过分别替换、 和 、和mallloc()来free()轻松printf()移植到ANSI C。newdeletestd::cout
如果要显示具有不同或更长字母表的元素，则将*alphabet参数更改为指向与 不同的字符串"abcdefg"。
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：*alphabet参数必须指向至少包含N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：在“7选4”中。
于 2020-12-11T02:00:38.880 回答
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
}
于 2020-12-12T10:21:51.297 回答
0
最近IronScripter网站上出现了一个需要 n-choose-k 解决方案的 PowerShell 挑战。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仅长度为 ChooseK 的组合，还是长度为 1 到 ChooseK 的组合。
- Prefix 参数实际上是输出字符串的累加器，但其效果是为初始调用传入的值实际上会为每一行输出添加前缀。
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
于 2021-04-14T01:12:02.577 回答
0
您可以使用 Asif 算法生成所有可能的组合。这可能是最简单和最有效的一种。你可以在这里查看中篇文章。
让我们看一下 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 ) );
于 2021-07-13T10:10:24.400 回答
-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;
}
这是我在 C++ 中的实现，它将所有组合写入指定的文件，但是可以更改行为，我生成了各种字典，它接受最小和最大长度和字符范围，目前仅支持 ansi，它足以满足我的需要
于 2012-09-16T12:53:19.387 回答
-1
我想介绍我的解决方案。没有递归调用，也没有嵌套循环next。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]));
}
}
}
以及用法示例：
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);
}
于 2016-12-27T11:27:04.553 回答
-1
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
IronScripter网站上最近发布了一个与此问题类似的挑战，您可以在其中找到我的链接和其他一些解决方案。
于 2021-04-23T03:31:38.023 回答