Interview question speed linked list and speed pointer

Tencent 1 interview question: how to quickly find the location of the length of a single linked list of intermediate nodes? The normal method is to go through it first and find the middle node of 2/length from scratch. The algorithm complexity is: O(3*n/2). The faster way is to use the principle of fast and slow Pointers.

Fast and slow linked list: using the idea of ruler, set two Pointers (1 fast and 1 slow) *serach and *mid, both pointing to the head node of the single linked list at the beginning. But the *search pointer moves twice as fast as *mid. When *search gets to the tail, mid gets right in the middle. The algorithm complexity is: O(n/2)

int GetMidNode(LinkList *L,int elem){
  LinkList *search,*mid;
  mid = search = L; // Point to the head 
  while (search->next != NULL){ // When the next node exists  
    if (search->next->next!=NULL) {// Check if the next node of the next is empty  
      search = search->next->next;
      mid = mid->next;
    } 
    else
      search = search->next;
  }
  elem = mid->data;
  return elem;
} 

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