Binary search cannot be efficiently applied to linked lists because it requires random access to theÂ elements.

Linked lists only support sequential access, which means you can't directly jump to the middle element like you can with arrays orÂ vectors.

However, you can still perform a search on a linked list, but it won't be as efficient as binaryÂ search.

Binary search works by dividing the search space in half repeatedly, which requires $O(1)$ time access to the middleÂ element.

Linked lists, however, have $O(n)$ time access to elements, as you must traverse from the head node to reach any specificÂ element.

For linked lists, a linear search is more suitable. Hereâ€™s an example of performing a linear search on a singly linkedÂ list:

```
#include <iostream>
// Define a node in the linked list
struct Node {
int data;
Node* next;
};
// Function to search for a value in the linked list
bool linearSearch(Node* head, int target) {
Node* current = head;
while (current != nullptr) {
if (current->data == target) {
return true;
}
current = current->next;
}
return false;
}
int main() {
// Create a linked list: 1 -> 2 -> 3 -> 4 -> 5
Node n1{1, nullptr},
n2{2, nullptr},
n3{3, nullptr},
n4{4, nullptr},
n5{5, nullptr};
n1.next = &n2;
n2.next = &n3;
n3.next = &n4;
n4.next = &n5;
int target = 3;
bool found = linearSearch(&n1, target);
std::cout << "The number " << target
<< (found ? " was" : " was not") << " found";
}
```

`The number 3 was found`

If you need faster search capabilities, consider theseÂ alternatives:

**Convert to Array**: Copy the linked list elements to an array, sort the array, and then perform binary search.**Skip List**: A skip list is a layered linked list that allows faster search by skipping over nodes, achieving $O(log n)$ time complexity for search operations.**Binary Search Tree (BST)**: Convert your linked list to a BST for efficient searching.

Hereâ€™s a simple example to convert a linked list to an array and then perform a binaryÂ search:

```
#include <algorithm>
#include <iostream>
#include <vector>
struct Node {
int data;
Node* next;
};
std::vector<int> linkedListToArray(Node* head) {
std::vector<int> arr;
Node* current = head;
while (current != nullptr) {
arr.push_back(current->data);
current = current->next;
}
return arr;
}
int main() {
Node n1{1, nullptr},
n2{2, nullptr},
n3{3, nullptr},
n4{4, nullptr},
n5{5, nullptr};
n1.next = &n2;
n2.next = &n3;
n3.next = &n4;
n4.next = &n5;
std::vector<int> array = linkedListToArray(&n1);
std::sort(array.begin(), array.end());
bool found = std::binary_search(
array.begin(), array.end(), 3);
std::cout << "The number 3 "
<< (found ? "was" : "was not") << " found";
}
```

`The number 3 was found`

While binary search is not suitable for linked lists due to their sequential access nature, you can use linear search or consider alternative data structures to achieve efficientÂ searching.

Answers to questions are automatically generated and may not have been reviewed.

This Question is from the Lesson:### Binary Search in C++

An introduction to the advantages of binary search, and how to use it with the C++ standard library algorithms `binary_search()`

, `lower_bound()`

, `upper_bound()`

, and `equal_range()`