C# 如何获取arraylist的所有组合？

我有一个字符串数组列表“abcde” 我想要一个方法返回另一个arraylist，其中包含C中给定arraylist（例如ab、ac、ad…）的所有可能组合#

有人知道一个简单的方法吗

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注意：长度2的所有可能组合，如果长度是可变的（可以更改），则会更好。

关于需要长度2组合的注释：

string s = "abcde";
var combinations = from c in s
                   from d in s.Remove(s.IndexOf(c), 1)
                   select new string(new[] { c, d });
foreach (var combination in combinations) {
    Console.WriteLine(combination);
}

响应任意长度的编辑：

static IEnumerable<string> GetCombinations(string s, int length) {
    Guard.Against<ArgumentNullException>(s == null);
    if (length > s.Length || length == 0) {
        return new[] { String.Empty };
    if (length == 1) {
        return s.Select(c => new string(new[] { c }));
    }
    return from c in s
           from combination in GetCombinations(
               s.Remove(s.IndexOf(c), 1),
               length - 1
           )
           select c + combination;
}

输出：

abc, abd, abe, acb, acd, ace, adb, adc, ade, aeb, aec, aed, bac, bad, bae, bca,
bcd, bce, bda, bdc, bde, bea, bec, bed, cab, cad, cae, cba, cbd, cbe, cda, cdb,
cde, cea, ceb, ced, dab, dac, dae, dba, dbc, dbe, dca, dcb, dce, dea, deb, dec,
eab, eac, ead, eba, ebc, ebd, eca, ecb, ecd, eda, edb, edc

aa, ab, ac, ad, ae, ba, bb, bc, bd, be, ca, cb, cc, cd, ce, da, db, dc, dd, de, ea, eb, ec, ed, ee

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下面是我的泛型函数，它可以返回T类型的所有组合：

static IEnumerable<IEnumerable<T>> GetCombinations<T>(IEnumerable<T> list, int length)
{
    if (length == 1) return list.Select(t => new T[] { t });

    return GetCombinations(list, length - 1)
        .SelectMany(t => list, (t1, t2) => t1.Concat(new T[] { t2 }));
}

输出：

abc, abd, abe, acb, acd, ace, adb, adc, ade, aeb, aec, aed, bac, bad, bae, bca,
bcd, bce, bda, bdc, bde, bea, bec, bed, cab, cad, cae, cba, cbd, cbe, cda, cdb,
cde, cea, ceb, ced, dab, dac, dae, dba, dbc, dbe, dca, dcb, dce, dea, deb, dec,
eab, eac, ead, eba, ebc, ebd, eca, ecb, ecd, eda, edb, edc

aa, ab, ac, ad, ae, ba, bb, bc, bd, be, ca, cb, cc, cd, ce, da, db, dc, dd, de, ea, eb, ec, ed, ee

已更新

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请参阅我对其他场景的回答，例如排列和k组合等。

仅使用数组和递归组合数组中的数字：

    static int n = 4;
    int[] baseArr = { 1, 2, 3, 4 };
    int[] LockNums;

    static void Main(string[] args)
    {
        int len = baseArr.Length;
        LockNums = new int[n];

        for (int i = 0; i < n; i++)
        {
            int num = baseArr[i];
            DoCombinations(num, baseArr, len);
            //for more than 4 numbers the print screen is too long if we need to check the result next line will help
            //Console.ReadLine(); 

        }
    }

    private void DoCombinations(int lockNum, int[] arr, int arrLen )        
    {
        int n1 = arr.Length;
        // next line shows the difference in length between the previous and its previous array
        int point = arrLen - n1; 
        LockNums[n - arr.Length] = lockNum;

        int[] tempArr = new int[arr.Length - 1];
        FillTempArr(lockNum, arr, tempArr);

        //next condition will print the last number from the current combination
        if (arr.Length == 1)
        {
            Console.Write(" {0}", lockNum);
            Console.WriteLine();
        }

        for (int i = 0; i < tempArr.Length; i++)
        {
            if ((point == 1) && (i != 0))
            {
                //without this code the program will fail to print the leading number of the next combination
                //and 'point' is the exact moment when this code has to be executed
                PrintFirstNums(baseArr.Length - n1);
            }
            Console.Write(" {0}", lockNum);
            int num1 = tempArr[i];
            DoCombinations(num1, tempArr, n1);
        }
    }

    private void PrintFirstNums(int missNums)
    {
        for (int i = 0; i < missNums; i++)
        {
            Console.Write(" {0}", LockNums[i]);
        }
    }

    private void FillTempArr(int lockN, int[] arr, int[] tempArr)
    {
        int idx = 0;
        foreach (int number in arr)
        {
            if (number != lockN)
            {
                tempArr[idx++] = number;
            }
        }
    }

    private void PrintResult(int[] arr)
    {
        foreach (int num in arr)
        {
            Console.Write(" {0}", num);
        }
    }

static int n=4；
int[]baseArr={1,2,3,4}；
int[]LockNums；
静态void Main（字符串[]参数）
{
int len=基准长度；
LockNums=新整数[n]；
对于（int i=0；i

您的意思是所有长度组合=2吗？所有可能的长度组合2除非您的目标是.NET 1.1，否则您应该使用列表。ArrayList已弃用。正在使用windows窗体application@Tergiver：否，

ArrayList

类未被弃用。它实际上已经过时了。您还应该执行var retComb=getcombines（s，2）.Distinct（）。其中（p=>p[0]>=p[p.Length]）；要删除多余的项，并反转项，例如

aabd

测试您的方法是重复排列而不是组合排列@HalilKaskavalci，请查看更多选项。现在看起来好多了。您好，这是我的数字组合解决方案。该解决方案仅使用阵列。我之所以写这篇文章，是因为我在任何地方都没有遇到使用阵列的解决方案。该代码是为快速演示而设置的，否则它应该适用于任何数组长度。