Question

Write a function, def merge(stack1:LinkedStack, stack2:LinkedStack)-> LinkedStack:

merge()function merges two sorted LinkedStacks into one.
Test case:
stack1: 1
3
4
5
7
Stack 2: 2
6
8
9
This Function should return: 1
2
3
4
5
6
7
8
9

Answer 1

Here is an example implementation of the `merge()` function in Python:

```python

def merge(stack1, stack2):

mergedStack = LinkedStack()

current1 = stack1.top

current2 = stack2.top

while current1 is not None and current2 is not None:

if current1.data <= current2.data:

mergedStack.push(current1.data)

current1 = current1.next

else:

mergedStack.push(current2.data)

current2 = current2.next

# Traverse through remaining elements in stack1

while current1 is not None:

mergedStack.push(current1.data)

current1 = current1.next

# Traverse through remaining elements in stack2

while current2 is not None:

mergedStack.push(current2.data)

current2 = current2.next

return mergedStack

```

Using the provided test case:

```

stack1: 1 -> 3 -> 4 -> 5 -> 7

stack2: 2 -> 6 -> 8 -> 9

```

The `merge(stack1, stack2)` function will return:

```

mergedStack: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9

```

Step-by-step explanation:

To merge two sorted LinkedStacks into one, we can follow these steps in the `merge()` function:

1. Create an empty LinkedStack, let's call it `mergedStack`, to store the merged elements.

2. Create two variables, `current1` and `current2`, and set them as the top nodes of `stack1` and `stack2`, respectively.

3. Loop through the two stacks while both `current1` and `current2` are not None:

• a. Compare the data of `current1` and `current2`. If `current1` is smaller or equal to `current2`, push `current1.data` into `mergedStack` and move `current1` to the next node in `stack1`.
• b. If `current2` is smaller than `current1`, push `current2.data` into `mergedStack` and move `current2` to the next node in `stack2`.

4. After the loop, there might be remaining elements in either `stack1` or `stack2`. If `current1` is not None, traverse through the remaining elements in `stack1` and push them into `mergedStack`. If `current2` is not None, traverse through the remaining elements in `stack2` and push them into `mergedStack`.

5. Finally, return the `mergedStack`.

The merged stack contains the elements from both `stack1` and `stack2` in sorted order.