Reverse Linked List

Reverse a singly linked list.

 

Thoughts:

1.Iterative Method.

Loop the linked list, set two pointers, one called "first" to always point to the head of the list, the other called "tail" to mark the currently looped element, each time we loop to that element, we switch the first and tail element to iteratively finish the reverse function.

 

Code:

    public ListNode reverseList(ListNode head) {

        ListNode first = null;

        ListNode tail = head;

              ListNode tmp;

        while(tail!=null){

  

        tmp=first;

        first = tail;

        tail=tail.next;

        first.next=tmp;

        }

        

        return first;

    }

 

2.Recursive Method

There's code in one reply that spells it out, but you might find it easier to start from the bottom up, by asking and answering tiny questions (this is the approach in The Little Lisper):

1. What is the reverse of null (the empty list)? null.
2. What is the reverse of a one element list? the element.
3. What is the reverse of an n element list? the reverse of the second element on followed by the first element.

Code:

public ListNode Reverse(ListNode list)

{

    if (list == null) return null; // first question



    if (list.next == null) return list; // second question



    // third question - in Lisp this is easy, but we don't have cons

    // so we grab the second element (which will be the last after we reverse it)



    ListNode secondElem = list.next;



    // bug fix - need to unlink list from the rest or you will get a cycle

    list.next = null;



    // then we reverse everything from the second element on

    ListNode reverseRest = Reverse(secondElem);



    // then we join the two lists

    secondElem.Next = list;



    return reverseRest;

}