下面的代码根据1,2,3 = 3,2,1 = 2,3,1的逻辑返回所有不同的组合,因此它只返回该组数字的1个实例.

但是,我想更改该逻辑,以便它返回所有数字集的所有实例.

为了实现这一目标,我需要对“GetPowerSet”下面的LINQ查询做些什么？

    public void GetPowersets()
    {
        List<int> ints = new List<int>()
        {
            1,2,2,3,3
        };

        var results = GetPowerSet(ints);


        List<String> combinations = new List<String>();
        foreach (var result in results)
        {
            StringBuilder sb = new StringBuilder();
            foreach (var intValue in result.OrderBy(x => x))
            {
                sb.Append(intValue + ",");
            }
            combinations.Add(sb.ToString());
        }

        string c1 = string.Join("|", combinations.ToArray()).Replace(",|", "|");
        //c1 = "|1|2|1,2|2|1,2|2,2|1,2,2|3|1,3|2,3|1,2,3|2,3|1,2,3|2,2,3|1,2,2,3|3|1,3|2,3|1,2,3|2,3|1,2,3|2,2,3|1,2,2,3|3,3|1,3,3|2,3,3|1,2,3,3|2,3,3|1,2,3,3|2,2,3,3|1,2,2,3,3,"

    }

    public IEnumerable<IEnumerable<int>> GetPowerSet(List<int> list)
    {
        return from m in Enumerable.Range(0, 1 << list.Count)
                     select
                         from i in Enumerable.Range(0, list.Count)
                         where (m & (1 << i)) != 0
                         select list[i];
    }

这是我想要达到的最终结果:(没有重复的组合行：duplicate = 3,2,1和3,2,1是相同的东西.但是1,2,3和3,2,1不是同样的事情,两者都应该在最终结果)

1
2
3
1,2
1,3
2,1
2,3
2,2
3,1
3,2
3,3
1,2,3
1,2,2
1,3,2
1,3,3
2,1,3
2,1,2
2,3,1
2,3,2
2,3,3
2,2,1
2,2,3
3,1,2
3,1,3
3,2,1
3,2,2
3,2,3
3,3,1
3,3,2
1,2,3,2
1,2,3,3
1,2,2,3
1,3,2,2
1,3,2,3
1,3,3,2
2,1,3,2
2,1,3,3
2,1,2,3
2,3,1,2
2,3,1,3
2,3,2,1
2,3,2,3
2,3,3,1
2,3,3,2
2,2,1,3
2,2,3,1
2,2,3,3
3,1,2,2
3,1,2,3
3,1,3,2
3,2,1,2
3,2,1,3
3,2,2,1
3,2,2,3
3,2,3,1
3,2,3,2
3,3,1,2
3,3,2,1
3,3,2,2
1,2,3,2,3
1,2,3,3,2
1,2,2,3,3
1,3,2,2,3
1,3,2,3,2
1,3,3,2,2
2,1,3,2,3
2,1,3,3,2
2,1,2,3,3
2,3,1,2,3
2,3,1,3,2
2,3,2,1,3
2,3,2,3,1
2,3,3,1,2
2,3,3,2,1
2,2,1,3,3
2,2,3,1,3
2,2,3,3,1
3,1,2,2,3
3,1,2,3,2
3,1,3,2,2
3,2,1,2,3
3,2,1,3,2
3,2,2,1,3
3,2,2,3,1
3,2,3,1,2
3,2,3,2,1
3,3,1,2,2
3,3,2,1,2
3,3,2,2,1

执行此操作的“foreach”方法,一旦数字集太大(我预期LINQ不应该出现此问题),往往会导致“Out of Memory Exception”.这是我想要的,返回我想要的结果集.但它很慢并且存在性能问题.我也愿意接受如何改善它的建议.

public List<List<int>> GetAllCombinationsOfAllSizes(List<int> ints)
{
    List<List<int>> returnResult = new List<List<int>>();

    var distinctInts = ints.Distinct().ToList();
    for (int j = 0; j < distinctInts.Count(); j++)
    {
        var number = distinctInts[j];

        var newList = new List<int>();
        newList.Add(number);
        returnResult.Add(newList);

        var listMinusOneObject = ints.Select(x => x).ToList();
        listMinusOneObject.Remove(listMinusOneObject.Where(x => x == number).First());

        if (listMinusOneObject.Count() > 0)
        {
            _GetAllCombinationsOfAllSizes(listMinusOneObject, newList, ref returnResult);
        }
    }

    return returnResult;
}
public void _GetAllCombinationsOfAllSizes(List<int> ints, List<int> growingList, ref List<List<int>> returnResult)
{
    var distinctInts = ints.Distinct().ToList();
    for (int j = 0; j < distinctInts.Count(); j++)
    {
        var number = distinctInts[j];

        var newList = growingList.ToList();
        newList.Add(number);
        returnResult.Add(newList);

        var listMinusOneObject = ints.Select(x => x).ToList();
        listMinusOneObject.Remove(listMinusOneObject.Where(x => x == number).First());

        if (listMinusOneObject.Count() > 0)
        {
            _GetAllCombinationsOfAllSizes(listMinusOneObject, newList, ref returnResult);
        }
    }

}

我正在寻找的答案是：如何实现我想要的结果集,但是使用LINQ和C#来实现它,其方式比我发布的当前“foreach”方式更快,更高效？

解决方法:

NEW UPDATE(删除旧代码,性能优于OP代码,产生输出)

static IEnumerable<int[]> EnumeratePermutations2(int[] ints)
{
    Dictionary<int, int> intCounts = ints.GroupBy(n => n)
                                         .ToDictionary(g => g.Key, g => g.Count());
    int[] distincts = intCounts.Keys.ToArray();
    foreach (int[] permutation in EnumeratePermutations2(new int[0], intCounts, distincts))
        yield return permutation;
}

static IEnumerable<int[]> EnumeratePermutations2(int[] prefix, Dictionary<int, int> intCounts, int[] distincts)
{
    foreach (int n in distincts)
    {
        int[] newPrefix = new int[prefix.Length + 1];
        Array.Copy(prefix, newPrefix, prefix.Length);
        newPrefix[prefix.Length] = n;
        yield return newPrefix;
        intCounts[n]--;
        int[] newDistincts = intCounts[n] > 0
                             ? distincts
                             : distincts.Where(x => x != n).ToArray();
        foreach (int[] permutation in EnumeratePermutations2(newPrefix, intCounts, newDistincts))
            yield return permutation;
        intCounts[n]++;
    }
}