How to use the two pointers trick in a linked list

A very simple trick which has multiple uses: finding the middle of a linked list, finding a certain node from the end (e.g. the 3rd node from the end back) or finding if a linked list has loops. I'll just post the code here, using the same structure as in the previous post. It has comments and it's really simple.

Finding the middle of a linked list:

   1:  int findMiddle(node* head) {

   2:      node* middle = head; // the middle will be here

   3:      while (head != 0 && head->next != 0) {

   4:          head = head->next->next; // advance the head by two positions forward

   5:          middle = middle->next; // advance the middle by only one position

   6:      }

   7:      return middle->data;

   8:  }

The solution is O(n) because the algorithm has to go through all the nodes once.

Finding if a linked list has loops (uses almost the same code):

   1:  bool hasLoops(node* head) {

   2:      node* slowPtr = head; //we'll move slower into the list using this pointer

   3:      while (head != 0 && head->next != 0) {

   4:          head = head->next->next; // advance the head by two positions forward

   5:          slowPtr = slowPtr->next; // advance the slowPtr by only one position

   6:          if (head == slowPtr) //these can never become equal unless one pointer loops

   7:              return true;     //back to a previous position, which means there's a loop

   8:      }

   9:      return false;

  10:  }

Again, the solution is O(n) in the worst case because the algorithm has to go through (potentially) all the nodes - it depends on where the loop is.