Returns the k-th node of the linked list.

There are a lot of ideas. The simplest thing is to record all the knots from the end of a traversal chain table to the end of the table, and calculate the number of the next to the last K from the beginning.

Of course, there are also technical skills.

With two pointers, ahead first goes through the K-1 step, then ahead and bhead walk together until a comes to the end, it is clear that B is the last K position.

Note:

1. Determine whether the input pointer is null;

2. Determine whether the K-1 is meaningful;

ReferenceCode:

Listnode * findk (listnode * List, unsigned int K) {If (list = NULL) return NULL; If (k <= 0) return NULL; listnode * ahead = List; listnode * bhead = List; For (unsigned int I = 0; I <K-1; I ++) {ahead = ahead-> next;} while (ahead->! = NULL) {bhead = bhead-> next; ahead = ahead-> next;} return bhead ;}