此递归函数运行后提示陷入无穷递归，不知道有关问题在何处，求指导

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;
  }
  }

[解决办法]
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;
}
}
[解决办法]
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;            }        }