Question

在python中使用一些代码按顺序输出字符串的特殊组合：如何在c＃

上做到这一点

def getOrderCombinations(me):
    l = len(me)
    looped = me+ me
    for start in range(0, l):
        for length in range(1, l):
            print(looped[start:start+length])
which gives:

>>> getOrderCombinations("ABCD")
A
AB
ABC
B
BC
BCD
C
CD
CDA
D
DA
DAB

我在尝试

public static string[] Combinations(string str)
{
    if (str.Length == 1)    
        return new string[] { str };
    char c = str[str.Length - 1];
    //here I was planning to make recursion 
    string[] returnArray = Combinations(str.Substring(0, str.Length - 1));
    // keep final string combinations
    List<string> finalArray = new List<string>();

    //how to loop correctly and avoid getting all combinations
    foreach (string s in returnArray)
        finalArray.Add(s);
    finalArray.Add(c.ToString());
}

对于字符串'ABCD'，输出应为'A'，'B'，'C'，'D'，'AB'，'BC'，'CD'，'DA'，'ABC'，'BCD '，'CDA'，DAB'。因此，字符串长度n的可能子串的数量将总是n *（n-1）。

如何在c＃上进行操作？

Answer 1

在python中实现它的方式有什么问题？直接端口应该可以很好地工作，我想：

public IEnumerable<string> GetOrderCombinations(string me)
{
  int    l      = me.Length ;
  string looped = me + me   ;

  for ( int start = 0 ; start < l ; ++start )
  {
    for ( int length = 1 ; length < l ; ++length )
    {
      yield return looped.Substring( start , length ) ;
    }
  }
}

你甚至可以把它变成Linq one-liner，几乎是Python实现的直接端口：

    public static IEnumerable<string> GetOrderCombinations( string me )
    {
        string looped = me + me;
        return Enumerable
               .Range(0,me.Length)
               .SelectMany( x => Enumerable.Range(1,me.Length) , looped.Substring )
               ;
    }

Answer 2

编辑：请考虑以下代码......

string str = "LEQN";
List<char> characters = str.ToCharArray().ToList();
List<string> combinations = new List<string>();
for (int i = 0; i < characters.Count; i++)
{
    int combCount = 1;
    string comb = characters[i].ToString();
    combinations.Add(comb);
    for (int j = 1; j < characters.Count - 1; j++)
    {
        int k = i + j;
        if (k >= characters.Count)
        {
            k = k - characters.Count;
        }
        comb += characters[k];
        combinations.Add(comb);
        combCount++;
    }
}

祝你好运！

以下是上述代码的输出......

enter image description here

Answer 3

这应该有效：致电 - ＆gt; GetCombinations（ “ABCD”）;

    List<string> allcombinations = new List<string>();

    public void GetCombinations(string input)
    {
        GetCombinations(1, input);
    }
    private void GetCombinations(int combLength, string input)
    {
        string current = string.Empty;
        for (int index = 0; index < input.Length; ++index)
        {
            if (index + combLength <= input.Length)
            {
                current = input.Substring(index, combLength);
                allcombinations.Add(current);
            }
            else
            {
                int leftOver = input.Length - index;
                current = input.Substring(index, leftOver);
                current += input.Substring(0, combLength - leftOver);
                allcombinations.Add(current);
            }
        }
        if (combLength < input.Length - 1)
        {
            GetCombinations(++combLength, input);
        }
    }

获取字符串的特殊组合

3 个答案: