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analysis/04-reverse-linked-list/exapmle/reverse-linked-list/README.md
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Reversed list:
5 -> 4 -> 3 -> 2 -> 1 -> null
```
## Memory walkthrough
see memory_walkthrough.md

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# Reverse Linked List — Memory Walkthrough
This walkthrough explains the classic three-pointer linked list reversal using a memory-oriented view.
The goal is not only to show that the algorithm works, but also to show what happens to:
- stack variables
- heap nodes
- `next` fields inside each node
The example list contains three nodes:
```text
1 -> 2 -> 3 -> null
```
For clarity, fake addresses are used:
```text
Node 1: 0x1000
Node 2: 0x2000
Node 3: 0x3000
```
These addresses are illustrative only.
A real program will use different addresses.
---
## Algorithm
```cpp
Node* reverseList(Node* head) {
Node* previous = nullptr;
Node* current = head;
while(current != nullptr) {
Node* next = current->next;
current->next = previous;
previous = current;
current = next;
}
return previous;
}
```
The three important pointers are:
| Pointer | Meaning |
|---|---|
| `previous` | Head of the already reversed part |
| `current` | Node currently being processed |
| `next` | Saved pointer to the remaining original list |
The most important rule is:
> Save `next` before changing `current->next`.
Otherwise the rest of the original list may become unreachable.
---
## Step 0 — Initial State
Before the loop starts:
```cpp
Node* previous = nullptr;
Node* current = head;
```
### Stack
```text
+----------+----------+
| Variable | Value |
+----------+----------+
| head | 0x1000 |
| previous | nullptr |
| current | 0x1000 |
| next | not set |
+----------+----------+
```
### Heap
```text
0x1000
+-----------+-----------+
| value = 1 | next=2000 |
+-----------+-----------+
0x2000
+-----------+-----------+
| value = 2 | next=3000 |
+-----------+-----------+
0x3000
+-----------+-----------+
| value = 3 | next=null |
+-----------+-----------+
```
### Logical view
```text
head/current
|
v
1 -> 2 -> 3 -> null
previous -> null
```
At this point, nothing has been reversed yet.
---
## Step 1 — Save `next`
Inside the first loop iteration:
```cpp
Node* next = current->next;
```
`current` points to Node 1.
`current->next` points to Node 2.
So:
```text
next = 0x2000
```
### Stack
```text
+----------+----------+
| Variable | Value |
+----------+----------+
| head | 0x1000 |
| previous | nullptr |
| current | 0x1000 |
| next | 0x2000 |
+----------+----------+
```
### Heap
```text
0x1000
+-----------+-----------+
| value = 1 | next=2000 |
+-----------+-----------+
0x2000
+-----------+-----------+
| value = 2 | next=3000 |
+-----------+-----------+
0x3000
+-----------+-----------+
| value = 3 | next=null |
+-----------+-----------+
```
Nothing in the heap changed yet.
The `next` stack variable only saves access to the rest of the list.
Without this temporary pointer, Node 2 and Node 3 could be lost after the next operation.
---
## Step 2 — Reverse `current->next`
Now the algorithm changes the link:
```cpp
current->next = previous;
```
`current` is Node 1.
`previous` is `nullptr`.
So Node 1 no longer points to Node 2.
It now points to `nullptr`.
### Stack
```text
+----------+----------+
| Variable | Value |
+----------+----------+
| head | 0x1000 |
| previous | nullptr |
| current | 0x1000 |
| next | 0x2000 |
+----------+----------+
```
### Heap
```text
0x1000
+-----------+-----------+
| value = 1 | next=null |
+-----------+-----------+
0x2000
+-----------+-----------+
| value = 2 | next=3000 |
+-----------+-----------+
0x3000
+-----------+-----------+
| value = 3 | next=null |
+-----------+-----------+
```
### Logical view
```text
current
|
v
1 -> null
next
|
v
2 -> 3 -> null
```
This is the key mutation.
The original list is now split into two logical parts:
```text
Reversed part:
1 -> null
Remaining original part:
2 -> 3 -> null
```
---
## Step 3 — Move `previous`
The algorithm advances the reversed part:
```cpp
previous = current;
```
`previous` now points to Node 1.
### Stack
```text
+----------+----------+
| Variable | Value |
+----------+----------+
| head | 0x1000 |
| previous | 0x1000 |
| current | 0x1000 |
| next | 0x2000 |
+----------+----------+
```
### Heap
```text
0x1000
+-----------+-----------+
| value = 1 | next=null |
+-----------+-----------+
0x2000
+-----------+-----------+
| value = 2 | next=3000 |
+-----------+-----------+
0x3000
+-----------+-----------+
| value = 3 | next=null |
+-----------+-----------+
```
### Logical view
```text
previous/current
|
v
1 -> null
next
|
v
2 -> 3 -> null
```
The reversed part now officially starts at `previous`.
---
## Step 4 — Move `current`
The algorithm continues with the saved next node:
```cpp
current = next;
```
`current` now points to Node 2.
### Stack
```text
+----------+----------+
| Variable | Value |
+----------+----------+
| head | 0x1000 |
| previous | 0x1000 |
| current | 0x2000 |
| next | 0x2000 |
+----------+----------+
```
### Heap
```text
0x1000
+-----------+-----------+
| value = 1 | next=null |
+-----------+-----------+
0x2000
+-----------+-----------+
| value = 2 | next=3000 |
+-----------+-----------+
0x3000
+-----------+-----------+
| value = 3 | next=null |
+-----------+-----------+
```
### Logical view
```text
previous
|
v
1 -> null
current
|
v
2 -> 3 -> null
```
The first iteration is complete.
---
## Step 5 — Second Iteration: Save `next`
The loop repeats.
```cpp
Node* next = current->next;
```
`current` points to Node 2.
Node 2 points to Node 3.
So:
```text
next = 0x3000
```
### Stack
```text
+----------+----------+
| Variable | Value |
+----------+----------+
| head | 0x1000 |
| previous | 0x1000 |
| current | 0x2000 |
| next | 0x3000 |
+----------+----------+
```
### Heap
```text
0x1000
+-----------+-----------+
| value = 1 | next=null |
+-----------+-----------+
0x2000
+-----------+-----------+
| value = 2 | next=3000 |
+-----------+-----------+
0x3000
+-----------+-----------+
| value = 3 | next=null |
+-----------+-----------+
```
Again, the heap has not changed yet.
---
## Step 6 — Second Iteration: Reverse Link
```cpp
current->next = previous;
```
`current` is Node 2.
`previous` is Node 1.
So Node 2 now points back to Node 1.
### Stack
```text
+----------+----------+
| Variable | Value |
+----------+----------+
| head | 0x1000 |
| previous | 0x1000 |
| current | 0x2000 |
| next | 0x3000 |
+----------+----------+
```
### Heap
```text
0x1000
+-----------+-----------+
| value = 1 | next=null |
+-----------+-----------+
0x2000
+-----------+-----------+
| value = 2 | next=1000 |
+-----------+-----------+
0x3000
+-----------+-----------+
| value = 3 | next=null |
+-----------+-----------+
```
### Logical view
```text
current
|
v
2 -> 1 -> null
next
|
v
3 -> null
```
The reversed part will become:
```text
2 -> 1 -> null
```
after `previous` moves to Node 2.
---
## Step 7 — Second Iteration: Move Pointers
```cpp
previous = current;
current = next;
```
### Stack
```text
+----------+----------+
| Variable | Value |
+----------+----------+
| head | 0x1000 |
| previous | 0x2000 |
| current | 0x3000 |
| next | 0x3000 |
+----------+----------+
```
### Heap
```text
0x1000
+-----------+-----------+
| value = 1 | next=null |
+-----------+-----------+
0x2000
+-----------+-----------+
| value = 2 | next=1000 |
+-----------+-----------+
0x3000
+-----------+-----------+
| value = 3 | next=null |
+-----------+-----------+
```
### Logical view
```text
previous
|
v
2 -> 1 -> null
current
|
v
3 -> null
```
Now two nodes are reversed.
---
## Step 8 — Third Iteration: Save `next`
```cpp
Node* next = current->next;
```
`current` is Node 3.
Node 3 points to `nullptr`.
So:
```text
next = nullptr
```
### Stack
```text
+----------+----------+
| Variable | Value |
+----------+----------+
| head | 0x1000 |
| previous | 0x2000 |
| current | 0x3000 |
| next | nullptr |
+----------+----------+
```
### Heap
```text
0x1000
+-----------+-----------+
| value = 1 | next=null |
+-----------+-----------+
0x2000
+-----------+-----------+
| value = 2 | next=1000 |
+-----------+-----------+
0x3000
+-----------+-----------+
| value = 3 | next=null |
+-----------+-----------+
```
---
## Step 9 — Third Iteration: Reverse Link
```cpp
current->next = previous;
```
`current` is Node 3.
`previous` is Node 2.
So Node 3 now points to Node 2.
### Stack
```text
+----------+----------+
| Variable | Value |
+----------+----------+
| head | 0x1000 |
| previous | 0x2000 |
| current | 0x3000 |
| next | nullptr |
+----------+----------+
```
### Heap
```text
0x1000
+-----------+-----------+
| value = 1 | next=null |
+-----------+-----------+
0x2000
+-----------+-----------+
| value = 2 | next=1000 |
+-----------+-----------+
0x3000
+-----------+-----------+
| value = 3 | next=2000 |
+-----------+-----------+
```
### Logical view
```text
current
|
v
3 -> 2 -> 1 -> null
```
The whole list is now reversed, but the loop still needs to update the stack pointers.
---
## Step 10 — Third Iteration: Move Pointers
```cpp
previous = current;
current = next;
```
Since `next` is `nullptr`, `current` becomes `nullptr`.
### Stack
```text
+----------+----------+
| Variable | Value |
+----------+----------+
| head | 0x1000 |
| previous | 0x3000 |
| current | nullptr |
| next | nullptr |
+----------+----------+
```
### Heap
```text
0x1000
+-----------+-----------+
| value = 1 | next=null |
+-----------+-----------+
0x2000
+-----------+-----------+
| value = 2 | next=1000 |
+-----------+-----------+
0x3000
+-----------+-----------+
| value = 3 | next=2000 |
+-----------+-----------+
```
### Logical view
```text
previous
|
v
3 -> 2 -> 1 -> null
current -> null
```
The loop condition fails:
```cpp
while(current != nullptr)
```
because `current` is now `nullptr`.
---
## Step 11 — Return New Head
At the end:
```cpp
return previous;
```
`previous` points to Node 3.
Node 3 is the new head of the reversed list.
### Final stack view
```text
+----------+----------+
| Variable | Value |
+----------+----------+
| old head | 0x1000 |
| new head | 0x3000 |
+----------+----------+
```
### Final heap view
```text
0x3000
+-----------+-----------+
| value = 3 | next=2000 |
+-----------+-----------+
0x2000
+-----------+-----------+
| value = 2 | next=1000 |
+-----------+-----------+
0x1000
+-----------+-----------+
| value = 1 | next=null |
+-----------+-----------+
```
### Final logical view
```text
new head
|
v
3 -> 2 -> 1 -> null
```
---
## Why the Temporary `next` Pointer Matters
This line is not optional:
```cpp
Node* next = current->next;
```
Without it, this operation:
```cpp
current->next = previous;
```
would overwrite the only pointer to the remaining original list.
For example, at the beginning:
```text
1 -> 2 -> 3 -> null
```
If Node 1 is changed to:
```text
1 -> null
```
before saving Node 2, then Node 2 and Node 3 are no longer reachable from any local variable.
That is why the algorithm always follows this order:
```cpp
Node* next = current->next; // preserve the remaining list
current->next = previous; // reverse the link
previous = current; // grow the reversed part
current = next; // continue with the remaining part
```
The order is the algorithm.
---
## Summary
During the algorithm:
- `previous` points to the already reversed part.
- `current` points to the node being processed.
- `next` preserves access to the not-yet-processed part.
- Only one `next` field is modified per iteration.
- No nodes are copied.
- No new list is allocated.
- The original nodes are relinked in-place.
The algorithm is small, but it is easy to get wrong because it mutates the structure while traversing it.
That is why a memory-level walkthrough is often more useful than just showing the final code.