problem report: middle of linked list

Middle of Linked List

propose: get the middle element of a linked list

method: 1. use two pointers

conplexity: o(n)

example:

Given 1->2->3, return the node with value 2.

Given 1->2, return the node with value 1.

mycode:

        ListNode firstPointer = new ListNode(0);
        ListNode secondPointer = new ListNode(0);
        if (head == null) {
            return null;
        }
        firstPointer = head;
        secondPointer = head.next;

        if (secondPointer == null || secondPointer.next == null) {
            return firstPointer;
        }

 if (head == null || head.next == null) {
            return head;
        }

        ListNode slow = head, fast = head.next;

reflections: 1. the third element euqal to null, do not need to consider as an special case! The two return statement of if actually do the same things. Use head to represend the return value. Use two different pointers as return value looks massy. (Reconsider the special case, tell whether there is an easy way to represent,especially for several if statement)

　　　　　　2. it‘s more reasonalbe to name the two pointer slow and fast

　　　　　　3. differece of stack and heap variable

iterative body part:

mycode:

while( secondPointer.next != null && secondPointer.next.next != null) {
            secondPointer = secondPointer.next.next;
            firstPointer = firstPointer.next;
        }

        if (secondPointer.next != null) {
            firstPointer = firstPointer.next;
        }
        return firstPointer;

answers:

1  while (fast != null && fast.next != null) {
2             slow = slow.next;
3             fast = fast.next.next;
4         }
5
6         return slow;

refelection: consider null element in the end of linked list as an normal node. Test fast and fast.next in the while loop can simplify the code.