Sub tasks

reorder can be achieved via

  1. cut from the middle into left and right
  2. reverse the right part
  3. join the linked list

Some O(n) method to deal with a linked list

Reverse a linked list

  1. walk through the linked list
  2. set node.next to prev
  3. store current node the prev
  4. loop

the initialization of prev is the tail NULL

Finding the middle of a linked list

fast-slow pointer.

  1. fast runner goes 2 steps each time
  2. slow runner goes 1 step
  3. when fast reach the end of the linked list, the slow id the middle of the list