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analysis/04-reverse-linked-list/diagrams/00_initial_state.png
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analysis/04-reverse-linked-list/diagrams/00_initial_state.puml
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@startuml
title Step 0 — Initial State
object "Node 1" as n1 {
value = 1
}
object "Node 2" as n2 {
value = 2
}
object "Node 3" as n3 {
value = 3
}
n1 --> n2 : next
n2 --> n3 : next
n3 --> "null" : next
object "previous" as prev
object "current" as cur
prev --> "null"
cur --> n1
note bottom
Before the loop starts:
previous = null
current = head
The original list is still intact.
end note
@enduml

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analysis/04-reverse-linked-list/diagrams/01_save_next.png
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analysis/04-reverse-linked-list/diagrams/01_save_next.puml
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@startuml
title Step 1 — Save next
object "Node 1" as n1 {
value = 1
}
object "Node 2" as n2 {
value = 2
}
object "Node 3" as n3 {
value = 3
}
n1 --> n2 : next
n2 --> n3 : next
n3 --> "null" : next
object "previous" as prev
object "current" as cur
object "next" as nxt
prev --> "null"
cur --> n1
nxt --> n2
note right of nxt
next = current->next
We save the next node before changing
current->next.
Without this temporary pointer, the rest
of the original list would become unreachable.
end note
@enduml

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analysis/04-reverse-linked-list/diagrams/02_reverse_current_link.png
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analysis/04-reverse-linked-list/diagrams/02_reverse_current_link.puml
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@startuml
title Step 2 — Reverse current->next
object "Node 1" as n1 {
value = 1
}
object "Node 2" as n2 {
value = 2
}
object "Node 3" as n3 {
value = 3
}
n1 --> "null" : next
n2 --> n3 : next
n3 --> "null" : next
object "previous" as prev
object "current" as cur
object "next" as nxt
prev --> "null"
cur --> n1
nxt --> n2
note right of n1
current->next = previous
Node 1 no longer points to Node 2.
It now points to the already reversed part.
At the first iteration, the reversed part is empty,
so Node 1 points to null.
end note
@enduml

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analysis/04-reverse-linked-list/diagrams/03_move_pointers.png
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analysis/04-reverse-linked-list/diagrams/03_move_pointers.puml
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@startuml
title Step 3 — Move previous and current
object "Node 1" as n1 {
value = 1
}
object "Node 2" as n2 {
value = 2
}
object "Node 3" as n3 {
value = 3
}
n1 --> "null" : next
n2 --> n3 : next
n3 --> "null" : next
object "previous" as prev
object "current" as cur
prev --> n1
cur --> n2
note right
previous = current
current = next
The reversed part is now:
1 -> null
The remaining original part is still:
2 -> 3 -> null
end note
@enduml

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analysis/04-reverse-linked-list/diagrams/04_second_iteration.png
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analysis/04-reverse-linked-list/diagrams/04_second_iteration.puml
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@startuml
title Step 4 — Second Iteration After Reversing Node 2
object "Node 1" as n1 {
value = 1
}
object "Node 2" as n2 {
value = 2
}
object "Node 3" as n3 {
value = 3
}
n2 --> n1 : next
n1 --> "null" : next
n3 --> "null" : next
object "previous" as prev
object "current" as cur
object "next" as nxt
prev --> n2
cur --> n3
nxt --> n3
note bottom
After processing Node 2:
2 -> 1 -> null
The reversed part grows from the front.
The remaining part starts at current.
end note
@enduml

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analysis/04-reverse-linked-list/diagrams/05_final_state.png
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analysis/04-reverse-linked-list/diagrams/05_final_state.puml
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@startuml
title Step 5 — Final State
object "Node 1" as n1 {
value = 1
}
object "Node 2" as n2 {
value = 2
}
object "Node 3" as n3 {
value = 3
}
n3 --> n2 : next
n2 --> n1 : next
n1 --> "null" : next
object "head" as head
object "previous" as prev
object "current" as cur
head --> n3
prev --> n3
cur --> "null"
note right of head
When current becomes null,
previous points to the new head.
head = previous
end note
@enduml

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analysis/04-reverse-linked-list/diagrams/README.md
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# Reverse Linked List — PlantUML Diagrams
This directory contains PlantUML diagrams for the three-pointer linked list reversal algorithm.
The diagrams use a small list:
```text
1 -> 2 -> 3 -> null
```
and show how it becomes:
```text
3 -> 2 -> 1 -> null
```
## Files
- `00_initial_state.puml` — initial state before the loop
- `01_save_next.puml` — saving `next = current->next`
- `02_reverse_current_link.puml` — reversing `current->next`
- `03_move_pointers.puml` — moving `previous` and `current`
- `04_second_iteration.puml` — state after the second node is processed
- `05_final_state.puml` — final state after the loop
## Generate PNG Files
```sh
plantuml diagrams/*.puml
```
## Generate SVG Files
```sh
plantuml -tsvg diagrams/*.puml
```
## Core Idea
During the loop, the list is logically split into two parts:
- `previous` points to the already reversed part
- `current` points to the node currently being processed
- `next` temporarily preserves access to the remaining original list
The key operation is:
```cpp
current->next = previous;
```
But this is only safe after saving:
```cpp
Node* next = current->next;
```
Otherwise the remaining part of the original list would be lost.

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analysis/04-reverse-linked-list/exapmle/reverse-linked-list/.gitignore vendored
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# This file is used to ignore files which are generated
# ----------------------------------------------------------------------------
*~
*.autosave
*.a
*.core
*.moc
*.o
*.obj
*.orig
*.rej
*.so
*.so.*
*_pch.h.cpp
*_resource.rc
*.qm
.#*
*.*#
core
!core/
tags
.DS_Store
.directory
*.debug
Makefile*
*.prl
*.app
moc_*.cpp
ui_*.h
qrc_*.cpp
Thumbs.db
*.res
*.rc
/.qmake.cache
/.qmake.stash
# qtcreator generated files
*.pro.user*
CMakeLists.txt.user*
# xemacs temporary files
*.flc
# Vim temporary files
.*.swp
# Visual Studio generated files
*.ib_pdb_index
*.idb
*.ilk
*.pdb
*.sln
*.suo
*.vcproj
*vcproj.*.*.user
*.ncb
*.sdf
*.opensdf
*.vcxproj
*vcxproj.*
# MinGW generated files
*.Debug
*.Release
# Python byte code
*.pyc
# Binaries
# --------
*.dll
*.exe

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analysis/04-reverse-linked-list/exapmle/reverse-linked-list/README.md
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# Reverse Linked List Example
This directory contains a small standalone C++ example for the classic three-pointer linked list reversal algorithm.
## Build
```bash
g++ -std=c++17 -Wall -Wextra -pedantic main.cpp -o reverse_linked_list
```
## Run
```bash
./reverse_linked_list
```
## Expected Output
```text
Original list:
1 -> 2 -> 3 -> 4 -> 5 -> null
Reversed list:
5 -> 4 -> 3 -> 2 -> 1 -> null
```
## Memory walkthrough
see memory_walkthrough.md

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analysis/04-reverse-linked-list/exapmle/reverse-linked-list/main.cpp
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/**
* @file main.cpp
* @brief Demonstrates the classic three-pointer algorithm for reversing a singly linked list.
*
* This example is intentionally small and self-contained.
* It is not meant to show that reversing linked lists is a common production task.
* Instead, it documents the interview pattern clearly enough that a reader unfamiliar
* with it can compile the program, run it, and inspect the output.
*/
#include <iostream>
#include <initializer_list>
/**
* @brief A minimal singly linked list node.
*
* Each node stores an integer value and a pointer to the next node.
* The last node in the list has @c next equal to @c nullptr.
*/
struct Node {
int value; ///< Payload stored in the node.
Node *next; ///< Pointer to the next node, or nullptr for the last node.
};
/**
* @brief Appends a new value to the end of the list.
*
* @param head Reference to the head pointer of the list.
* @param value Value to store in the new node.
*
* This helper is used only to build the demonstration list.
* It keeps the example simple and avoids using STL containers for the list itself,
* because the goal is to demonstrate raw pointer manipulation.
*/
void appendNode (Node *&head, int value) {
Node *node = new Node{value, nullptr};
if (head == nullptr) {
head = node;
return;
}
Node *current = head;
while (current->next != nullptr)
current = current->next;
current->next = node;
}
/**
* @brief Creates a linked list from an initializer list.
*
* @param values Values to insert into the list in the given order.
* @return Pointer to the first node of the created list.
*
* The caller owns the returned list and must release it with freeList().
*/
Node *createList (std::initializer_list<int> values) {
Node *head = nullptr;
for (int value : values)
appendNode (head, value);
return head;
}
/**
* @brief Prints the list without modifying it.
*
* @param head Pointer to the first node of the list.
*
* Output example:
* @code
* 1 -> 2 -> 3 -> 4 -> null
* @endcode
*/
void printList (const Node *head) {
const Node *current = head;
while (current != nullptr) {
std::cout << current->value << " -> ";
current = current->next;
}
std::cout << "null" << std::endl;
}
/**
* @brief Reverses a singly linked list in place.
*
* @param head Pointer to the first node of the original list.
* @return Pointer to the first node of the reversed list.
*
* This is the classic three-pointer interview algorithm.
*
* The three pointers are:
*
* - @c previous — the already reversed part of the list
* - @c current — the node we are processing right now
* - @c next — the original next node saved before we overwrite @c current->next
*
* Why @c next is necessary:
*
* In a singly linked list, each node only knows where the next node is.
* When we execute:
*
* @code
* current->next = previous;
* @endcode
*
* we destroy the original forward link.
* Without saving it first, the rest of the list would be lost.
*
* The algorithm works by moving one node at a time from the original forward chain
* into the reversed chain.
*
* Initial state:
*
* @code
* previous = null
* current = 1 -> 2 -> 3 -> 4 -> null
* @endcode
*
* After processing node 1:
*
* @code
* previous = 1 -> null
* current = 2 -> 3 -> 4 -> null
* @endcode
*
* After processing node 2:
*
* @code
* previous = 2 -> 1 -> null
* current = 3 -> 4 -> null
* @endcode
*
* When @c current becomes @c nullptr, @c previous points to the new head.
*
* Complexity:
*
* - Time: O(n), because each node is visited once.
* - Extra memory: O(1), because only a fixed number of pointers is used.
*/
Node *reverseList (Node *head) {
Node *previous = nullptr;
Node *current = head;
while (current != nullptr) {
/*
* Save the original next node before changing current->next.
* Without this line, the rest of the list would become unreachable.
*/
Node *next = current->next;
/*
* Reverse the direction of the link.
* The current node now points to the already reversed part.
*/
current->next = previous;
/*
* Move previous forward.
* The current node becomes the new head of the reversed part.
*/
previous = current;
/*
* Continue with the node that originally followed current.
*/
current = next;
}
return previous;
}
/**
* @brief Releases all nodes in the list.
*
* @param head Pointer to the first node of the list.
*
* This function is separated from the reversal and printing logic.
* It exists only because this example uses raw @c new to keep the node structure explicit.
*/
void freeList (Node *head) {
Node *current = head;
while (current != nullptr) {
Node *next = current->next;
delete current;
current = next;
}
}
/**
* @brief Program entry point.
*
* Builds a small list, prints it, reverses it, prints it again,
* and finally releases all allocated nodes.
*/
int main() {
Node *list = createList ({1, 2, 3, 4, 5});
std::cout << "Original list:" << std::endl;
printList (list);
list = reverseList (list);
std::cout << "\nReversed list:" << std::endl;
printList (list);
freeList (list);
return 0;
}

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analysis/04-reverse-linked-list/exapmle/reverse-linked-list/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.

5
analysis/04-reverse-linked-list/exapmle/reverse-linked-list/output_example.txt
View File

@@ -1,5 +0,0 @@
Original list:
1 -> 2 -> 3 -> 4 -> 5 -> null
Reversed list:
5 -> 4 -> 3 -> 2 -> 1 -> null

7
analysis/04-reverse-linked-list/exapmle/reverse-linked-list/reverse-linked-list.pro
View File

@@ -1,7 +0,0 @@
TEMPLATE = app
CONFIG += console c++17
CONFIG -= app_bundle
CONFIG -= qt
SOURCES += \
main.cpp

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analysis/04-reverse-linked-list/readme.md
View File

@@ -1,255 +0,0 @@
# Analysis #04 — Reverse Linked List: Academic Exercise
## Problem
A classic interview question:
Given a singly linked list:
```cpp
struct Node {
int value;
Node* next;
};
```
Reverse the list:
```text
1 -> 2 -> 3 -> 4 -> null
```
into:
```text
4 -> 3 -> 2 -> 1 -> null
```
using:
* O(n) time
* O(1) additional memory
---
## Typical Interview Solution
The standard solution uses three pointers:
```cpp
Node* previous = nullptr;
Node* current = head;
while(current) {
Node* next = current->next;
current->next = previous;
previous = current;
current = next;
}
head = previous;
```
The candidate is expected to produce this solution quickly and correctly.
---
## What This Actually Tests
Despite its popularity, this problem tests a surprisingly narrow set of skills.
Primarily:
* Pointer manipulation
* Attention to detail
* Familiarity with linked lists
* Prior exposure to a common interview pattern
In many cases, prior exposure matters more than reasoning.
A candidate who has seen the problem twenty times may solve it in under a minute.
A senior engineer with years of production experience may need significantly longer if they have never encountered this specific exercise before.
---
## Why This Is Rare In Real Engineering
The interesting question is:
> When was the last time you actually reversed a linked list in production code?
For most engineers, the answer is:
> Almost never.
Modern systems rarely use linked lists as a primary data structure.
More commonly you will encounter:
* vectors
* deques
* ring buffers
* hash tables
* trees
* databases
* message queues
The embedded world is similar.
Typical structures include:
* circular buffers
* DMA buffers
* message queues
* routing tables
* state machines
Linked lists certainly exist.
However, fully reversing one is rarely a real business requirement.
---
## The Hidden Assumption
The interview question starts with an assumption:
> You already have a linked list.
Real engineering often starts with a different question:
> Why is this a linked list in the first place?
That decision is usually far more important than the reversal algorithm itself.
---
## Real-World Equivalent
Finding a true production equivalent is difficult.
Most real systems solve a different problem.
### Example 1: Event History Viewer
A user wants to see the newest events first.
A typical engineering solution is:
* iterate in reverse
* change presentation logic
* adjust query ordering
The underlying data structure often remains unchanged.
---
### Example 2: CAN Trace Analysis
Suppose a trace contains millions of CAN frames.
The user wants the newest messages displayed at the top.
Nobody reverses the entire dataset.
Instead:
* reverse iteration is used
* the UI changes presentation order
* indexing structures provide efficient access
The stored data remains exactly as it was.
---
## What Real Engineers Usually Ask
A more practical engineering question would be:
> The user wants to view data in reverse order.
>
> Do we actually need to modify the data structure?
This question frequently leads to better solutions.
---
## Better Interview Question
Instead of asking:
> Reverse a linked list.
Consider asking:
> A system stores ten million records.
>
> Users want to view them in reverse order.
>
> What solution options exist, and what are their trade-offs?
Now the discussion becomes much more interesting:
* memory usage
* cache locality
* ownership
* indexing
* performance
* maintainability
* user requirements
In other words:
Engineering begins.
---
## Common Mistakes
* ❌ Assuming data must be modified to change presentation order
* ❌ Ignoring alternative data structures
* ❌ Focusing on implementation before understanding requirements
* ❌ Treating algorithmic manipulation as the only valid solution
---
## Key Takeaway
Reverse Linked List is useful as an educational exercise.
It teaches pointer manipulation and careful reasoning about memory.
However, the problem itself rarely appears in production software in its original form.
The real engineering question is usually not:
> How do we reverse the list?
Instead it is:
> Do we need to reverse it at all?
---
## Project Perspective
> Exists in real engineering?
**Partially**
Linked lists exist.
Complete list reversal is a very uncommon business requirement.
> Exists in interview form?
**Yes**
It remains one of the most common classic coding interview questions.
The exercise is valuable for learning pointer manipulation.
Its usefulness as a predictor of engineering ability is far less obvious.