sorted list of Doubly linked list. - doubly-linked-list

Question

please help me with the insert method please. i want to insert an int which have a value greater than that int in the doubly linked list. then the new int will replace the greater value from the doubly linked list (i got that working fine). but now i want that greater int to go down the list and be inserted where appropriate.
if i say insert 2 4 6 8 , it will work (result: 2 4 6 8)
if i then insert 1 (which is smaller than 2)
result will be 1 4 6 8
2
the problem is, when i insert lets say 7 (greater than 6, but smaller than 8) i dont know how to go down row and insert where appropriate.
enter code here
public void insert(int newInt) {
Cell newCell = new Cell (newInt);
if (this.size() == 0){
smallest = newCell;
}
for (Cell i = smallest; i != null; i = i.right){
// if newCell is the larger than all in row, then add it to the end.
if (newCell.value > i.value && i.right == null){
i.right = newCell;
}
// if newCell is smaller than smallest, then add it to smallest and old smallest will go down the list (compare and put it where appropriate).
if (newCell.value < smallest.value){
newCell.below = smallest;
newCell.right = smallest.right;
smallest.right = null;
smallest = newCell;
}
// if newCell is smallest than any element, then replace that cell with newCell, and send that cell to below row
// do again untill all below rows are sorted. hence this could be a while loop.
if (newCell.value > i.value && newCell.value < i.right.value){
System.out.println("newCell is " + newCell);
Cell largerRight = i.right;
newCell.right = largerRight.right;
largerRight.right = null;
i.right = newCell;
temp = smallest.below;
while (temp != null){
System.out.println(temp);
for (Cell k = temp; k != null; k = k.right){
if (largerRight.value > temp.value && temp.right == null){
temp.right = largerRight;
temp = null;
}
}
temp =temp.below;
}
}
}
}//end insert

Answer 1

When inserting into a doubly linked list, you almost always have a first and last pointer. You also have a new node that you want inserted in the list, but as yet without the pointer information.
I'll do this in pseudo-code so you can translate it to whatever language you want (and since it may be homework).
First, detect an empty list. If that's the case, just build a new one-element list:
if first == NULL:
newNode->next = NULL # one element pointing nowhere.
newNode->prev = NULL
first = newNode # first and last are that element.
last = newNode
return
Otherwise, find the first node greater than the current, that's where you want to insert before:
insertPoint = first # scan list until end.
while insertPoint != NULL:
if insertPoint->payload > newNode->payload: # stop if found one greater.
break;
insertPoint = insertPoint->next # move to next.
Now, there are three distinct possibilities here:
insertPoint is the first node, meaning you have to insert at the front;
insertPoint is NULL, meaning you have to append to the list; or
insertPoint is a valid node other than the first.
First, handle the two special cases. These are considered special since they change the control information of the list (first and last), rather than just pointers within the nodes themselves:
if insertPoint == first: # inserting at front.
newNode->next = first # new node points forwards to current first
newNode->prev = NULL # and backwards to nowhere.
first = newNode # and update first.
return
if insertPoint = NULL: # appending to end.
newNode->next = NULL # new node points forwards to nowhere
newNode->prev = last # and backwards to current last.
last = newNode # and update last.
return
If you've gotten to here, you simply need to insert, knowing that you have a valid current and previous node:
newNode->next = insertPoint # set new node to point to correct nodes.
newNode->prev = insertPoint->prev
insertPoint->prev->next = newNode # update those nodes to point to new node.
insertPoint->prev = newNode
return