递归Java功能，有例外

Question

我有一个带有IDS＆amp;的表他们的邻居需要创建一个递归函数，该功能从开始ID到结束ID >，而无需两次越过同一点。假设启动ID为1，端ID为3。唯一可能的完整路径是(1,2,3(＆amp;(1,5,3(

{1 | 2,5}
{2 | 1,3,4,5}
{3 | 2,5}
{4 | 2}
{5 | 1,2,3}

当前代码(通过@Jeffrey Phillips Freeman(

List<Integer> searchHops(int from, int to, List<Integer> seen) {
    seen.add(from);
    if (from == to)
        return new ArrayList<Integer>(Arrays.asList(from));
    for (int neighbor : getNeighbors(from))
        if (!seen.contains(neighbor)) {
            List<Integer> result = searchHops(neighbor, to, seen);
            if (result != null) {
                result.add(0, from);
                return result;
            }
        }
    return null;
}

Answer 1

沿着以下内容的某些内容应该可以解决问题。我将留给您实施getneighbors方法。您还应该调用该方法，看到的是一个空的arraylist

private List<Integer> searchHops(int from, int to, List<Integer> seen) {
  seen.add(from);
  if( from == to )
    return new ArrayList<Integer>(Arrays.asList(from));
  for(int neighbor : getNeighbors(from) )
    if( !seen.contains(neighbor) ) {
      List<Integer> result = searchHops(neighbor, to, seen);
      if(result != null) {
        result.add(0, from);
        return result;
      }
    }
  return null;
}
//actual entry point
public List<Integer> searchHops(int from, int to) {
    return searchHops(from, to, new ArrayList<Integer>());
}