我发现这个线程很有用，我想我会添加一个Javascript解决方案，你可以弹出到Firebug。取决于你的JS引擎，如果起始字符串很大，可能会花一点时间。

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");

输出如下:

abc
ab
ac
a
bc
b
c

我发现这个线程很有用，我想我会添加一个Javascript解决方案，你可以弹出到Firebug。取决于你的JS引擎，如果起始字符串很大，可能会花一点时间。

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");

输出如下:

abc
ab
ac
a
bc
b
c

我的实现在c/c++

#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;
}

这里我的实现在c++，它写所有的组合到指定的文件，但行为可以改变，我在生成各种字典，它接受最小和最大长度和字符范围，目前只有ANSI支持，它足以满足我的需求

在VB。Net，该算法从一组数字(PoolArray)中收集n个数字的所有组合。例如，从“8,10,20,33,41,44,47”中选择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)

像Andrea Ambu一样用Python写的，但不是硬编码来选择三个。

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

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

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

            return seq

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

我已经编写了一个类来处理处理二项式系数的常见函数，这是您的问题属于的问题类型。它执行以下任务:

Outputs all the K-indexes in a nice format for any N choose K to a file. The K-indexes can be substituted with more descriptive strings or letters. This method makes solving this type of problem quite trivial. Converts the K-indexes to the proper index of an entry in the sorted binomial coefficient table. This technique is much faster than older published techniques that rely on iteration. It does this by using a mathematical property inherent in Pascal's Triangle. My paper talks about this. I believe I am the first to discover and publish this technique, but I could be wrong. Converts the index in a sorted binomial coefficient table to the corresponding K-indexes. Uses Mark Dominus method to calculate the binomial coefficient, which is much less likely to overflow and works with larger numbers. The class is written in .NET C# and provides a way to manage the objects related to the problem (if any) by using a generic list. The constructor of this class takes a bool value called InitTable that when true will create a generic list to hold the objects to be managed. If this value is false, then it will not create the table. The table does not need to be created in order to perform the 4 above methods. Accessor methods are provided to access the table. There is an associated test class which shows how to use the class and its methods. It has been extensively tested with 2 cases and there are no known bugs.

要了解这个类并下载代码，请参见将二项式系数表化。

将这个类转换为c++应该不难。