假设我有一个包含单个字符的任意大小的数组。我想计算这些字符的所有可能组合，直到任意长度。
所以我要说我的数组是[1,2,3]。用户指定的长度为2.然后可能的组合为[11,22,33,12,13,23,21,31,32]。
我很难找到一个允许任意长度的合适算法，而不仅仅是排列数组。哦，虽然速度并非绝对关键，但它也应该相当快。
答案 0 :(得分：9)
只需携带即可添加。
假设您的数组包含4个符号，并且您需要长度为3的符号。
以000开头（即您的单词上的每个符号=字母[0]）
然后加起来：
000 001 002 003 010 011 ...
算法（给定这些指数）只是为了增加最低数量。如果它达到字母表中的符号数，请增加之前的数字（遵循相同的规则）并将当前设置为0.
C ++代码：
int N_LETTERS = 4;
char alphabet[] = {'a', 'b', 'c', 'd'};
std::vector<std::string> get_all_words(int length)
{
std::vector<int> index(length, 0);
std::vector<std::string> words;
while(true)
{
std::string word(length);
for (int i = 0; i < length; ++i)
word[i] = alphabet[index[i]];
words.push_back(word);
for (int i = length-1; ; --i)
{
if (i < 0) return words;
index[i]++;
if (index[i] == N_LETTERS)
index[i] = 0;
else
break;
}
}
}
代码未经测试，但应该可以解决问题。
答案 1 :(得分：2)
一种方法是使用一个简单的计数器，在内部将其解释为基数N，其中N是数组中的项数。然后，您从基本N计数器中提取每个数字，并将其用作数组的索引。因此，如果您的数组是[1,2]并且用户指定的长度是2，那么您有
Counter = 0, indexes are 0, 0
Counter = 1, indexes are 0, 1
Counter = 2, indexes are 1, 0
Counter = 3, indexes are 1, 1
这里的技巧将是你的base-10到base-N转换代码，这并不是非常困难。
答案 2 :(得分：2)
Knuth在The Art of Computer Programming，第1卷中涵盖了某种深度的组合和排列。这是我几年前写的一种算法的实现（不要讨厌它的风格，它的古代代码）：< / p>
#include <algorithm>
#include <vector>
#include <functional>
#include <iostream>
using namespace std;
template<class BidirectionalIterator, class Function, class Size>
Function _permute(BidirectionalIterator first, BidirectionalIterator last, Size k, Function f, Size n, Size level)
{
// This algorithm is adapted from Donald Knuth,
// "The Art of Computer Programming, vol. 1, p. 45, Method 1"
// Thanks, Donald.
for( Size x = 0; x < (n-level); ++x ) // rotate every possible value in to this level's slot
{
if( (level+1) < k )
// if not at max level, recurse down to twirl higher levels first
f = _permute(first,last,k,f,n,level+1);
else
{
// we are at highest level, this is a unique permutation
BidirectionalIterator permEnd = first;
advance(permEnd, k);
f(first,permEnd);
}
// rotate next element in to this level's position & continue
BidirectionalIterator rotbegin(first);
advance(rotbegin,level);
BidirectionalIterator rotmid(rotbegin);
rotmid++;
rotate(rotbegin,rotmid,last);
}
return f;
}
template<class BidirectionalIterator, class Function, class Size>
Function for_each_permutation(BidirectionalIterator first, BidirectionalIterator last, Size k, Function fn)
{
return _permute<BidirectionalIterator,Function,Size>(first, last, k, fn, distance(first,last), 0);
}
template<class Elem>
struct DumpPermutation : public std::binary_function<bool, Elem* , Elem*>
{
bool operator()(Elem* begin, Elem* end) const
{
cout << "[";
copy(begin, end, ostream_iterator<Elem>(cout, " "));
cout << "]" << endl;
return true;
}
};
int main()
{
int ary[] = {1, 2, 3};
const size_t arySize = sizeof(ary)/sizeof(ary[0]);
for_each_permutation(&ary[0], &ary[arySize], 2, DumpPermutation<int>());
return 0;
}
该计划的输出是：
[1 2 ]
[1 3 ]
[2 3 ]
[2 1 ]
[3 1 ]
[3 2 ]
如果您希望组合包含重复的元素，如[11] [22]和[33]，您可以使用上面的算法生成组合列表，然后通过执行类似的操作将附加到生成的列表中的新元素这样：
for( size_t i = 0; i < arySize; ++i )
{
cout << "[";
for( int j = 0; j < k; ++j )
cout << ary[i] << " ";
cout << "]" << endl;
}
...现在程序输出变为：
[1 2 ]
[1 3 ]
[2 3 ]
[2 1 ]
[3 1 ]
[3 2 ]
[1 1 ]
[2 2 ]
[3 3 ]
答案 3 :(得分：1)
如果你事先知道长度，你需要的只是一些for循环。比如说，长度= 3：
for ( i = 0; i < N; i++ )
for ( j = 0; j < N; j++ )
for ( k = 0; k < N; k++ )
you now have ( i, j, k ), or a_i, a_j, a_k
现在概括一下，只需递归执行，递归的每一步都有一个for循环：
recurse( int[] a, int[] result, int index)
if ( index == N ) base case, process result
else
for ( i = 0; i < N; i++ ) {
result[index] = a[i]
recurse( a, result, index + 1 )
}
当然，如果您只想要所有组合，则可以将每个步骤视为基于N的数字，从1到k^N - 1，其中k是长度。
基本上你会得到，在N基础上（k = 4）：
0000 // take the first element four times
0001 // take the first element three times, then the second element
0002
...
000(N-1) // take the first element three times, then take the N-th element
1000 // take the second element, then the first element three times
1001
..
(N-1)(N-1)(N-1)(N-1) // take the last element four times
答案 4 :(得分：0)
使用Peter的算法效果很好;但是，如果您的字母集太大或字符串大小太长，尝试将所有排列放在数组中并返回数组将无法正常工作。数组的大小将是字符串的大小，增加到字符串的长度。
我在perl中创建了这个以解决问题：
package Combiner;
#package used to grab all possible combinations of a set of letters. Gets one every call, allowing reduced memory usage and faster processing.
use strict;
use warnings;
#initiate to use nextWord
#arguments are an array reference for the list of letters and the number of characters to be in the generated strings.
sub new {
my ($class, $phoneList,$length) = @_;
my $self = bless {
phoneList => $phoneList,
length => $length,
N_LETTERS => scalar @$phoneList,
}, $class;
$self->init;
$self;
}
sub init {
my ($self) = shift;
$self->{lindex} = [(0) x $self->{length}];
$self->{end} = 0;
$self;
}
#returns all possible combinations of N phonemes, one at a time.
sub nextWord {
my $self = shift;
return 0 if $self->{end} == 1;
my $word = [('-') x $self->{length}];
$$word[$_] = ${$self->{phoneList}}[${$self->{lindex}}[$_]]
for(0..$self->{length}-1);
#treat the string like addition; loop through 000, 001, 002, 010, 020, etc.
for(my $i = $self->{length}-1;;$i--){
if($i < 0){
$self->{end} = 1;
return $word;
}
${$self->{lindex}}[$i]++;
if (${$self->{lindex}}[$i] == $self->{N_LETTERS}){
${$self->{lindex}}[$i] = 0;
}
else{
return $word;
}
}
}
这样称呼：my $c = Combiner->new(['a','b','c','d'],20);。然后拨打nextWord以获取下一个字词;如果nextWord返回0，则表示已完成。
答案 5 :(得分：0)
这是Haskell的my implementation：
g :: [a] -> [[a]] -> [[a]]
g alphabet = concat . map (\xs -> [ xs ++ [s] | s <- alphabet])
allwords :: [a] -> [[a]]
allwords alphabet = concat $ iterate (g alphabet) [[]]
将此脚本加载到GHCi。假设我们想要在字母表中找到长度小于或等于2的所有字符串{＆＃39; a＆＃39;＆＃39; b＆＃39;＆＃39; c＆＃39;}。以下GHCi会话就是这样做的：
*Main> take 13 $ allwords ['a','b','c']
["","a","b","c","aa","ab","ac","ba","bb","bc","ca","cb","cc"]
或者，如果你只想要长度等于2的字符串：
*Main> filter (\xs -> length xs == 2) $ take 13 $ allwords ['a','b','c']
["aa","ab","ac","ba","bb","bc","ca","cb","cc"]
小心allwords ['a','b','c']因为它是一个无限的列表！
答案 6 :(得分：0)
这是我写的。可能对你有帮助...
#include<stdio.h>
#include <unistd.h>
void main()
{
FILE *file;
int i=0,f,l1,l2,l3=0;
char set[]="abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ01234567890!@#$%&*.!@#$%^&*()";
int size=sizeof(set)-1;
char per[]="000";
//check urs all entered details here//
printf("Setlength=%d Comination are genrating\n",size);
// writing permutation here for length of 3//
for(l1=0;l1<size;l1++)
//first for loop which control left most char printed in file//
{
per[0]=set[l1];
// second for loop which control all intermediate char printed in file//
for(l2=0;l2<size;l2++)
{
per[1]=set[l2];
//third for loop which control right most char printed in file//
for(l3=0;l3<size;l3++)
{
per[2]=set[l3];
//apend file (add text to a file or create a file if it does not exist.//
file = fopen("file.txt","a+");
//writes array per to file named file.txt//
fprintf(file,"%s\n",per);
///Writing to file is completed//
fclose(file);
i++;
printf("Genrating Combination %d\r",i);
fflush(stdout);``
usleep(1);
}
}
}
printf("\n%d combination has been genrate out of entered data of length %d \n",i,size);
puts("No combination is left :) ");
puts("Press any butoon to exit");
getchar();
}