/// <summary>
/// Split a list by index
/// Original list will end at the index
/// The rest of list will be return as a new list
/// </summary>
/// <param name="list"></param>
/// <param name="i"></param>
/// <returns></returns>
public static LList <T> Split(LList <T> list, int i)
{
if (list.IsEmpty() || list.Count() == 1)
{
return(null);
}
LList <T> result = new LList <T>();
var temp = list.FindByPosition(i);
if (temp == null)
{
return(null);
}
result.First = temp.Next;
if (i == list.Count() - 1)
{
result.Last = temp.Next;
}
else
{
result.Last = list.Last;
}
temp.Next = null;
return(result);
}
/// <summary>
/// Reverse strategy 2
/// traverse from (first,last) , (first+1,last-1) ... swap each other's value
/// </summary>
public static void Reverse2(LList <T> list)
{
if (list.IsEmpty())
{
return;
}
T temp;
for (int i = 0; i < list.Count() / 2; i++)
{
temp = list.FindByPosition(i).Value;
list.FindByPosition(i).Value = list.FindByPosition(list.Count() - 1 - i).Value;
list.FindByPosition(list.Count() - 1 - i).Value = temp;
}
}
/// <summary>
/// reverse strategy 3;
/// change each node's pointer from "point to next" to "point to prev";
/// it needs to declare 3 pointers for different purpose;
/// a "firstHolder" to hold the first item;
/// a "current" to point to the node whose "Next" needs to redirect;
/// a "pointerHolder" to point to the node next to the current node, so once current redirect, this node still can be accessed;
/// see as a "a b c" sliding window slide through the list;
/// once "b" finishes redirection, whole window slide one node;
/// </summary>
/// <param name="list"></param>
/// <returns></returns>
public static void Reverse3(LList <T> list)
{
if (list.IsEmpty() || list.Count() == 1)
{
return;
}
var firstHolder = list.First;
var current = list.First.Next;
LListNode <T> pointerHolder;
while (current != null)
{
pointerHolder = current.Next;
current.Next = list.First;
list.First = current;
current = pointerHolder;
}
firstHolder.Next = null;
list.Last = firstHolder;
}
