在链接列表的最后一个元素中找到第k个

Question

I was trying to implement this recursively, but I'm not sure why this code doesn't work (this is assuming that I have a length function that returns correctly):

Node findk(Node head, int k) {
    if (node_length(head)==k) {
        return head; }
    else {
        return findk(head.next, k-1);}}

Thanks!

Answer 1

There are two issues with your code:

• you should not be decrementing k as you go down the list, and
• you should pay attention to hitting null before reaching k -th element when the list is too short.

Here is one possible fix:

Node findk(Node head, int k) {
    if (head == null) return null;
    if (node_length(head)==k) return head;
    return findk(head.next, k);
}

Note that this solution is O(n ² ), because node_length , which must be O(1), is called for each of Nk nodes of the list. There are several ways of doing it faster - for example, by finding int m = node_length(head) , and then returning (mk) -th node from the beginning of the list.

Answer 2

If you want to find the K'th element from the end You are decrementing the value of K which is wrong, the correct code is as below :

Node findk(Node head, int k) {
  if (node_length(head)==k) {
    return head; 
  } else {
    return findk(head.next, k);
  }
}

Also I hope your node_length() method takes care of the scenario where the Node passed to it is null.