// time complexity (O(M x N))
// space complexity (O(1)) because The iterative approach only allocates
// a few pointers, so it has a constant overall memory footprint.
public static ListNodeLc21 MergeTwoListV2(ListNodeLc21 list1, ListNodeLc21 list2)
{
//ListNodeLc21 dummyHead = new ListNodeLc21(-1);
//var prev = dummyHead;
//while (list1 != null && list2 != null)
//{
// if (list1.Value <= list2.Value)
// {
// prev.Next = list1;
// list1 = list1.Next;
// }
// else
// {
// prev.Next = list2;
// list2 = list2.Next;
// }
// prev = prev.Next;
//}
//prev.Next = list1 == null ? list2 : list1;
//return prev.Next;
ListNodeLc21 prehead = new ListNodeLc21(-1);
ListNodeLc21 prev = prehead;
while (list1 != null && list2 != null)
{
if (list1.Value <= list2.Value)
{
prev.Next = list1;
list1 = list1.Next;
}
else
{
prev.Next = list2;
list2 = list2.Next;
}
prev = prev.Next;
}
// exactly one of list1 and list2 can be non-null at this point, so connect
// the non-null list to the end of the merged list.
prev.Next = list1 ?? list2;
return(prehead.Next);
}
//Merge two sorted linked lists and return it as a new list.The new list should be made by splicing together the nodes of the first two lists.
// Example:
//Input: 1->2->4, 1->3->4
//Output: 1->1->2->3->4->4
// time complexity (O(M x N))
// space complexity (O(M x N)) because the both recursive call need to reach the end
public static ListNodeLc21 MergeTwoList(ListNodeLc21 list1, ListNodeLc21 list2)
{
if (list1 == null)
{
return(list2);
}
if (list2 == null)
{
return(list1);
}
if (list1.Value < list2.Value)
{
list1.Next = MergeTwoList(list1.Next, list2);
return(list1);
}
list2.Next = MergeTwoList(list1, list2.Next);
return(list2);
}
public ListNodeLc21(int Value, ListNodeLc21 node = null)
{
this.Value = Value;
Next = node;
}
