此递归函数运行后提示陷入无穷递归，不知道问题在何处，求指导。
public int fk(int i, List<int>s)
  {
  int min = 999;
  if (s.Count <=0)
  return Convert.ToInt32(d[0, i]);
  else
  {
  List<int> newt = new List<int>();
   
  for (int k = 0; k < s.Count; k++)
  {
  s.ForEach(z => newt.Add(z));
  int backup = newt[k];
  newt.RemoveAt(k);
  int value = fk(backup, newt);
  if (min > value + Convert.ToInt32(d[backup, i]))
  min = value + Convert.ToInt32(d[backup,i]);

  }
  return min;
  }
  }

------解决方案--------------------
public int fk(int i, List<int>s)
{
int min = 999;
if (s.Count <=0)
return Convert.ToInt32(d[0, i]);
else
{
List<int> newt = new List<int>();
s.ForEach(z => newt.Add(z));
for (int k = 0; k < s.Count; k++)
{
int backup = newt[k];
newt.RemoveAt(k);
int value = fk(backup, newt);
if (min > value + Convert.ToInt32(d[backup, i]))
min = value + Convert.ToInt32(d[backup,i]);

}
return min;
}
}
------解决方案--------------------
我找出问题所在了，哈哈，刚才那个只保证跳出循环了，我知道你的意思假如原始为{1，2}，先求{2}，在求{2}，最后得到值。

C# code

 public int fk(int i, List<int> s)
        {
            

            int min = 999;
            if (s.Count <= 0)
            {
                return Convert.ToInt32(d[0, i]);           
            }
            else
            {               
                for (int k = 0; k < s.Count; k++)
                {
                    List<int> newt = new List<int>();
                    s.ForEach(z => newt.Add(z));
                    int backup = newt[k];
                    newt.RemoveAt(k);                   
                    int value = fk(backup, newt);
                    if (min > value + Convert.ToInt32(d[backup, i]))
                        min = value + Convert.ToInt32(d[backup, i]);
                }               
                return min;
            }
        }